Question
A conductor with an inner cavity, shown in figure below, carries a total charge of plus 14 mu C. Within the cavity, insulated from the conductor, is a point charge of negative 10 mu C.q13.png
How much charge is on the outer surface of the conductor?
plus 4 mu C
plus 10 mu C
negative 24 mu C
None of the above
plus 14 mu C
How much charge is on the outer surface of the conductor?
plus 4 mu C
plus 10 mu C
negative 24 mu C
None of the above
plus 14 mu C
Answers
Answer
The figure below shows an arrangement of two electrons each 2 cross times 10 to the power of negative 10 end exponent m from a proton. What is the magnitude of the net electric force they will exert on the proton? q subscript e equals negative 1.6 cross times 10 to the power of negative 19 space end exponent C; q subscript p equals plus 1.6 cross times 10 to the power of negative 19 space end exponent C
q4.png
q4.png
Answer
The figure below shows an arrangement of two electrons each 2 cross times 10 to the power of negative 10 end exponent m from a proton. What is the magnitude of the net electric force they will exert on the proton? q subscript e equals negative 1.6 cross times 10 to the power of negative 19 space end exponent C; q subscript p equals plus 1.6 cross times 10 to the power of negative 19 space end exponent C
q4.png
8.63 cross times 10 to the power of negative 9 space end exponent N
4.98 cross times 10 to the power of negative 9 space end exponent N
9.29 cross times 10 to the power of negative 5 space end exponent N
None of the above
2.87 cross times 10 to the power of negative 9 space end exponent N
q4.png
8.63 cross times 10 to the power of negative 9 space end exponent N
4.98 cross times 10 to the power of negative 9 space end exponent N
9.29 cross times 10 to the power of negative 5 space end exponent N
None of the above
2.87 cross times 10 to the power of negative 9 space end exponent N
Answer
The work done to bring a unit positive charge from infinity to a point in an electric field is referred to as:
Electric dipole moment
Electric field intensity
None of the above
The total energy of the point charge
Electric potential
Electric dipole moment
Electric field intensity
None of the above
The total energy of the point charge
Electric potential
Answer
The electric potential at points in xy plane is given by V left parenthesis x comma y right parenthesis equals open parentheses 4 fraction numerator text v end text over denominator m squared end fraction close parentheses x squared minus open parentheses 9 fraction numerator text v end text over denominator m squared end fraction close parentheses y squared
What is the magnitude of the electric field at the point open parentheses 1 half semicolon 1 over 6 close parentheses?
E equals 5 space fraction numerator text v end text over denominator m end fraction
E equals 0.56 space fraction numerator text v end text over denominator m end fraction
None of the above
E equals 25 space fraction numerator text v end text over denominator m end fraction
E equals 0.75 space fraction numerator text v end text over denominator m end fraction
What is the magnitude of the electric field at the point open parentheses 1 half semicolon 1 over 6 close parentheses?
E equals 5 space fraction numerator text v end text over denominator m end fraction
E equals 0.56 space fraction numerator text v end text over denominator m end fraction
None of the above
E equals 25 space fraction numerator text v end text over denominator m end fraction
E equals 0.75 space fraction numerator text v end text over denominator m end fraction
Answer
A flat sheet of paper of area 0.40 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 15 N/C. Find the magnitude of the electric flux through the sheet.
text 3 N.m end text squared divided by C
text 6 N.m end text squared divided by C
text 5.2 N.m end text squared divided by C
None of the above
text 0 N.m end text squared divided by C
text 3 N.m end text squared divided by C
text 6 N.m end text squared divided by C
text 5.2 N.m end text squared divided by C
None of the above
text 0 N.m end text squared divided by C
Answer
A charge Q is enclosed by a gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will
increase by a factor
be twice
be reduced to half
None of the above
be four times more
increase by a factor
be twice
be reduced to half
None of the above
be four times more
Answer
Calculate the magnitude of the electric field at a point P which is 30 cm to the right of a point charge Q equals negative 3 cross times 10 to the power of negative 6 end exponent space text C end text
Answer
Calculate the magnitude of the electric field at a point P which is 30 cm to the right of a point charge Q equals negative 3 cross times 10 to the power of negative 6 end exponent space text C end text
30 space text N end text divided by text C end text
None of the above
9 cross times 10 to the power of 16 space text N end text divided by text C end text
3 cross times 10 to the power of 5 space text N end text divided by text C end text
9 cross times 10 squared space text N end text divided by text C end text
30 space text N end text divided by text C end text
None of the above
9 cross times 10 to the power of 16 space text N end text divided by text C end text
3 cross times 10 to the power of 5 space text N end text divided by text C end text
9 cross times 10 squared space text N end text divided by text C end text
Answer
Calculate the magnitude of the electric field at a point P which is 30 cm to the right of a point charge Q equals negative 3 cross times 10 to the power of negative 6 end exponent space text C end text
30 space text N end text divided by text C end text
None of the above
9 cross times 10 to the power of 16 space text N end text divided by text C end text
3 cross times 10 to the power of 5 space text N end text divided by text C end text
9 cross times 10 squared space text N end text divided by text C end text
30 space text N end text divided by text C end text
None of the above
9 cross times 10 to the power of 16 space text N end text divided by text C end text
3 cross times 10 to the power of 5 space text N end text divided by text C end text
9 cross times 10 squared space text N end text divided by text C end text
Answer
The density of charges that are spread on a thin copper wire of length l is called,
Answer
The density of charges that are spread on a thin copper wire of length l is called,
All of the above
Surface charge density
None of the above
Volume charge density
Linear charge density
All of the above
Surface charge density
None of the above
Volume charge density
Linear charge density
Answer
Electric charge is distributed uniformly along an infinitely long, thin wire as shown in figure below. The linear charge density λ is 1.5 cross times 10 to the power of negative 10 end exponent space C divided by m.Find the magnitude of electric field at 108 cm from the infinitely long wire.
Answer
The figure below shows three charges at the corners of an isoscleles triangle. The plus 3 mu C and negative 3 mu C charges form a dipole. Calculate the force negative 8 mu C charge exerts on the dipole.
q5.png
q5.png
Answered by
GPT 3.5
The total charge on the conductor includes both the charge within the cavity and the charge on the outer surface.
The charge within the cavity is negative 10 μC.
The total charge on the conductor is plus 14 μC.
Therefore, the charge on the outer surface of the conductor must be the difference between the total charge and the charge within the cavity.
Charge on the outer surface = Total charge - Charge within the cavity
= 14 μC - (-10 μC)
= 14 μC + 10 μC
= 24 μC
So, the charge on the outer surface of the conductor is 24 μC.
The correct answer is negative 24 μC.
The charge within the cavity is negative 10 μC.
The total charge on the conductor is plus 14 μC.
Therefore, the charge on the outer surface of the conductor must be the difference between the total charge and the charge within the cavity.
Charge on the outer surface = Total charge - Charge within the cavity
= 14 μC - (-10 μC)
= 14 μC + 10 μC
= 24 μC
So, the charge on the outer surface of the conductor is 24 μC.
The correct answer is negative 24 μC.
Answered by
GPT 3.5
The magnitude of the electric force between two charges can be calculated using Coulomb's Law:
F = k * |q1| * |q2| / r^2
where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.
In this case, both charges are electrons (negative 1.6 x 10^-19 C) and the proton is positive (1.6 x 10^-19 C). The distance between them is given as 2 x 10^-10 m.
Plugging in the values:
F = (8.99 x 10^9 N m^2 / C^2) * |(-1.6 x 10^-19 C)| * |(-1.6 x 10^-19 C)| / (2 x 10^-10 m)^2
F = (8.99 x 10^9 N m^2 / C^2) * (1.6 x 10^-19 C) * (1.6 x 10^-19 C) / (2 x 10^-10 m)^2
F ≈ 5.75 x 10^-8 N
Therefore, the magnitude of the net electric force that the two electrons will exert on the proton is approximately 5.75 x 10^-8 N.
F = k * |q1| * |q2| / r^2
where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.
In this case, both charges are electrons (negative 1.6 x 10^-19 C) and the proton is positive (1.6 x 10^-19 C). The distance between them is given as 2 x 10^-10 m.
Plugging in the values:
F = (8.99 x 10^9 N m^2 / C^2) * |(-1.6 x 10^-19 C)| * |(-1.6 x 10^-19 C)| / (2 x 10^-10 m)^2
F = (8.99 x 10^9 N m^2 / C^2) * (1.6 x 10^-19 C) * (1.6 x 10^-19 C) / (2 x 10^-10 m)^2
F ≈ 5.75 x 10^-8 N
Therefore, the magnitude of the net electric force that the two electrons will exert on the proton is approximately 5.75 x 10^-8 N.
Answered by
GPT 3.5
To find the magnitude of the net electric force between the two electrons and the proton, we can use Coulomb's Law:
F = k * |q_e| * |q_p| / r^2
where F is the force, k is the electrostatic constant, q_e and q_p are the charges of the electrons and proton respectively, and r is the distance between them.
Plugging in the values:
F = (8.99 x 10^9 N m^2 / C^2) * |(-1.6 x 10^-19 C)| * |(1.6 x 10^-19 C)| / (2 x 10^-10 m)^2
F = (8.99 x 10^9 N m^2 / C^2) * (1.6 x 10^-19 C) * (1.6 x 10^-19 C) / (2 x 10^-10 m)^2
F ≈ 1.837 x 10^-8 N
Therefore, the magnitude of the net electric force that the two electrons will exert on the proton is approximately 1.837 x 10^-8 N.
None of the given options match this value.
F = k * |q_e| * |q_p| / r^2
where F is the force, k is the electrostatic constant, q_e and q_p are the charges of the electrons and proton respectively, and r is the distance between them.
Plugging in the values:
F = (8.99 x 10^9 N m^2 / C^2) * |(-1.6 x 10^-19 C)| * |(1.6 x 10^-19 C)| / (2 x 10^-10 m)^2
F = (8.99 x 10^9 N m^2 / C^2) * (1.6 x 10^-19 C) * (1.6 x 10^-19 C) / (2 x 10^-10 m)^2
F ≈ 1.837 x 10^-8 N
Therefore, the magnitude of the net electric force that the two electrons will exert on the proton is approximately 1.837 x 10^-8 N.
None of the given options match this value.
Answered by
GPT 3.5
The work done to bring a unit positive charge from infinity to a point in an electric field is referred to as electric potential.
Answered by
GPT 3.5
To find the magnitude of the electric field at a point:
E = - ∇V
where E is the electric field, ∇ is the gradient operator, and V is the electric potential.
Taking the partial derivatives of V with respect to x and y:
∂V/∂x = 8v/m^2 * x
∂V/∂y = -18v/m^2 * y
Substituting the given coordinates (1/2, 1/6) into the partial derivatives:
∂V/∂x = 8v/m^2 * (1/2) = 4v/m^2
∂V/∂y = -18v/m^2 * (1/6) = -3v/m^2
Therefore, the electric field at the point (1/2, 1/6) is:
E = - (∂V/∂x, ∂V/∂y) = - (4v/m^2, -3v/m^2) = (-4v/m^2, 3v/m^2)
The magnitude of the electric field is:
|E| = sqrt((-4v/m^2)^2 + (3v/m^2)^2) = sqrt(16v^2/m^4 + 9v^2/m^4) = sqrt(25v^2/m^4) = 5v/m^2
Therefore, the magnitude of the electric field at the point (1/2, 1/6) is 5v/m.
The correct answer is E equals 5 fraction v over m.
E = - ∇V
where E is the electric field, ∇ is the gradient operator, and V is the electric potential.
Taking the partial derivatives of V with respect to x and y:
∂V/∂x = 8v/m^2 * x
∂V/∂y = -18v/m^2 * y
Substituting the given coordinates (1/2, 1/6) into the partial derivatives:
∂V/∂x = 8v/m^2 * (1/2) = 4v/m^2
∂V/∂y = -18v/m^2 * (1/6) = -3v/m^2
Therefore, the electric field at the point (1/2, 1/6) is:
E = - (∂V/∂x, ∂V/∂y) = - (4v/m^2, -3v/m^2) = (-4v/m^2, 3v/m^2)
The magnitude of the electric field is:
|E| = sqrt((-4v/m^2)^2 + (3v/m^2)^2) = sqrt(16v^2/m^4 + 9v^2/m^4) = sqrt(25v^2/m^4) = 5v/m^2
Therefore, the magnitude of the electric field at the point (1/2, 1/6) is 5v/m.
The correct answer is E equals 5 fraction v over m.
Answered by
GPT 3.5
The magnitude of the electric flux through a surface is given by:
Φ = E * A * cos(θ)
where Φ is the electric flux, E is the magnitude of the electric field, A is the area of the surface, and θ is the angle between the normal to the surface and the electric field.
In this case, the magnitude of the electric field is 15 N/C, the area of the sheet is 0.40 m^2, and the angle between the normal to the sheet and the electric field is 60°.
Plugging in the values:
Φ = (15 N/C) * (0.40 m^2) * cos(60°)
Φ ≈ (15 N/C) * (0.40 m^2) * 0.5
Φ ≈ 6 N.m^2/C
Therefore, the magnitude of the electric flux through the sheet is 6 N.m^2/C.
The correct answer is 6 N.m^2/C.
Φ = E * A * cos(θ)
where Φ is the electric flux, E is the magnitude of the electric field, A is the area of the surface, and θ is the angle between the normal to the surface and the electric field.
In this case, the magnitude of the electric field is 15 N/C, the area of the sheet is 0.40 m^2, and the angle between the normal to the sheet and the electric field is 60°.
Plugging in the values:
Φ = (15 N/C) * (0.40 m^2) * cos(60°)
Φ ≈ (15 N/C) * (0.40 m^2) * 0.5
Φ ≈ 6 N.m^2/C
Therefore, the magnitude of the electric flux through the sheet is 6 N.m^2/C.
The correct answer is 6 N.m^2/C.
Answered by
GPT 3.5
According to Gauss' law, the electric flux through a closed surface is directly proportional to the charge enclosed by that surface. The electric flux, Φ, is given by:
Φ = Q / ε₀
where Q is the charge enclosed by the surface, and ε₀ is the electric constant (also known as the permittivity of free space).
If the radius of the gaussian spherical surface is doubled, the surface area of the sphere increases by a factor of 4 (since the area of a sphere is proportional to the square of its radius). However, the charge enclosed by the surface remains the same.
As a result, the electric flux through the larger surface would be:
Φ' = Q / ε₀
Since Q/ε₀ is constant, the electric flux through the larger surface is the same as the electric flux through the original surface.
Therefore, the outward electric flux remains the same.
The correct answer is None of the above.
Φ = Q / ε₀
where Q is the charge enclosed by the surface, and ε₀ is the electric constant (also known as the permittivity of free space).
If the radius of the gaussian spherical surface is doubled, the surface area of the sphere increases by a factor of 4 (since the area of a sphere is proportional to the square of its radius). However, the charge enclosed by the surface remains the same.
As a result, the electric flux through the larger surface would be:
Φ' = Q / ε₀
Since Q/ε₀ is constant, the electric flux through the larger surface is the same as the electric flux through the original surface.
Therefore, the outward electric flux remains the same.
The correct answer is None of the above.
Answered by
GPT 3.5
To calculate the magnitude of the electric field at a point P due to a point charge Q, we can use the equation:
E = k * |Q| / r^2
where E is the electric field, k is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), |Q| is the absolute value of the charge, and r is the distance between the charge and the point P.
In this case, the charge Q is negative, -3 x 10^-6 C, and the distance from the charge to point P is 30 cm, or 0.3 m.
Plugging in the values:
E = (8.99 x 10^9 N m^2 / C^2) * |-3 x 10^-6 C| / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / 0.09 m^2
E ≈ 8.97 x 10^1 N/C
Therefore, the magnitude of the electric field at point P is approximately 8.97 x 10^1 N/C, or 89.7 N/C.
E = k * |Q| / r^2
where E is the electric field, k is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), |Q| is the absolute value of the charge, and r is the distance between the charge and the point P.
In this case, the charge Q is negative, -3 x 10^-6 C, and the distance from the charge to point P is 30 cm, or 0.3 m.
Plugging in the values:
E = (8.99 x 10^9 N m^2 / C^2) * |-3 x 10^-6 C| / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / 0.09 m^2
E ≈ 8.97 x 10^1 N/C
Therefore, the magnitude of the electric field at point P is approximately 8.97 x 10^1 N/C, or 89.7 N/C.
Answered by
GPT 3.5
To calculate the magnitude of the electric field at point P due to a point charge Q, we can use the equation:
E = k * |Q| / r^2
where E is the electric field, k is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), |Q| is the absolute value of the charge, and r is the distance between the charge and the point P.
In this case, the charge Q is negative, -3 x 10^-6 C, and the distance from the charge to point P is 30 cm, or 0.3 m.
Plugging in the values:
E = (8.99 x 10^9 N m^2 / C^2) * |(-3 x 10^-6 C)| / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / 0.09 m^2
E ≈ 9 x 10^5 N/C
Therefore, the magnitude of the electric field at point P is approximately 9 x 10^5 N/C, or 900,000 N/C.
The correct answer is 9 x 10^5 N/C.
E = k * |Q| / r^2
where E is the electric field, k is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), |Q| is the absolute value of the charge, and r is the distance between the charge and the point P.
In this case, the charge Q is negative, -3 x 10^-6 C, and the distance from the charge to point P is 30 cm, or 0.3 m.
Plugging in the values:
E = (8.99 x 10^9 N m^2 / C^2) * |(-3 x 10^-6 C)| / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / 0.09 m^2
E ≈ 9 x 10^5 N/C
Therefore, the magnitude of the electric field at point P is approximately 9 x 10^5 N/C, or 900,000 N/C.
The correct answer is 9 x 10^5 N/C.
Answered by
GPT 3.5
To calculate the magnitude of the electric field due to a point charge Q at a distance r from the charge, we use the formula:
E = k * |Q| / r^2
where E is the electric field, k is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), |Q| is the absolute value of the charge, and r is the distance between the charge and the point P.
In this case, the charge Q is given as -3 x 10^-6 C, and the distance r is 30 cm, or 0.3 m.
Plugging in the values:
E = (8.99 x 10^9 N m^2 / C^2) * |(-3 x 10^-6 C)| / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / 0.09 m^2
E ≈ 9 x 10^5 N/C
Therefore, the magnitude of the electric field at point P is approximately 9 x 10^5 N/C.
The correct answer is 9 cross times 10 to the power of 5 space text N end text divided by text C end text.
E = k * |Q| / r^2
where E is the electric field, k is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), |Q| is the absolute value of the charge, and r is the distance between the charge and the point P.
In this case, the charge Q is given as -3 x 10^-6 C, and the distance r is 30 cm, or 0.3 m.
Plugging in the values:
E = (8.99 x 10^9 N m^2 / C^2) * |(-3 x 10^-6 C)| / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / (0.3 m)^2
E = (8.99 x 10^9 N m^2 / C^2) * (3 x 10^-6 C) / 0.09 m^2
E ≈ 9 x 10^5 N/C
Therefore, the magnitude of the electric field at point P is approximately 9 x 10^5 N/C.
The correct answer is 9 cross times 10 to the power of 5 space text N end text divided by text C end text.
Answered by
GPT 3.5
The density of charges that are spread on a thin copper wire of length l is called charge density.
Answered by
GPT 3.5
The density of charges that are spread on a thin copper wire of length l is called linear charge density.
Answered by
GPT 3.5
To find the magnitude of the electric field at a distance r from an infinitely long, thin wire with linear charge density λ, we can use the formula:
E = (k * λ) / r
where E is the electric field, k is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), λ is the linear charge density, and r is the distance from the wire.
In this case, the linear charge density λ is given as 1.5 x 10^-10 C/m, and the distance from the wire is 108 cm, or 1.08 m.
Plugging in the values:
E = (8.99 x 10^9 N m^2 / C^2) * (1.5 x 10^-10 C/m) / 1.08 m
E = (8.99 x 10^9 N m^2 / C^2) * (1.5 x 10^-10 C) / 1.08
E ≈ 1.24 x 10^-1 N/C
Therefore, the magnitude of the electric field at a distance of 108 cm from the infinitely long wire is approximately 1.24 x 10^-1 N/C.
The correct answer is 1.24 x 10^-1
E = (k * λ) / r
where E is the electric field, k is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), λ is the linear charge density, and r is the distance from the wire.
In this case, the linear charge density λ is given as 1.5 x 10^-10 C/m, and the distance from the wire is 108 cm, or 1.08 m.
Plugging in the values:
E = (8.99 x 10^9 N m^2 / C^2) * (1.5 x 10^-10 C/m) / 1.08 m
E = (8.99 x 10^9 N m^2 / C^2) * (1.5 x 10^-10 C) / 1.08
E ≈ 1.24 x 10^-1 N/C
Therefore, the magnitude of the electric field at a distance of 108 cm from the infinitely long wire is approximately 1.24 x 10^-1 N/C.
The correct answer is 1.24 x 10^-1
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