The torque experienced by an electric dipole in a uniform electric field can be calculated using the formula:
τ = pE sinθ
where:
τ = torque
p = dipole moment (p = qd)
E = magnitude of the electric field
θ = angle between the dipole moment vector and the electric field vector
In this case, the dipole moment is given by p = qd, where q is the charge and d is the separation distance between the charges. We are given the values of q and d, so we can calculate the dipole moment:
p = (1.6 x 10^-19 C)(0.12 m) = 1.92 x 10^-20 C*m
We are also given the magnitude of the electric field, which is 10^5 N/C.
Now, let's calculate the angle θ. Since the electric field is directed parallel to the plane of the figure, the dipole moment vector will be perpendicular to the electric field vector, hence θ = 90 degrees.
sinθ = sin 90° = 1
Finally, let's calculate the torque:
τ = (1.92 x 10^-20 C*m)(10^5 N/C)(1) = 1.92 x 10^-15 N*m
Therefore, the magnitude of the torque is 1.92 x 10^-15 Nm.
Figure below shows an electric dipole in a uniform electric field with magnitude 10 cross times 10 to the power of 5 N/C directed parallel to the plane of the figure.The electric dipole consists of two charges q subscript 1 equals plus e and q subscript 2 equals negative e space left parenthesis e equals 1.6 cross times 10 to the power of negative 19 end exponent right parenthesis, which both lie in the plane and separated by a distance 0.12 space n text m end text
q6.png
Calculate the magnitude of the torque
1.92 cross times 10 to the power of negative 29 end exponent space text Nm end text
None of the above
1.23 cross times 10 to the power of negative 23 end exponent space text Nm end text
0 space text Nm end text
1.32 cross times 10 to the power of negative 33 end exponent space text Nm end text
5 answers
Figure below shows an electric dipole in a uniform electric field with magnitude 10 cross times 10 to the power of 5 N/C directed parallel to the plane of the figure.The electric dipole consists of two charges q subscript 1 equals plus e and q subscript 2 equals negative e space left parenthesis e equals 1.6 cross times 10 to the power of negative 19 end exponent right parenthesis, which both lie in the plane and separated by a distance 0.12 space n text m end text
q6.png
Calculate the magnitude of the torque
1.92 cross times 10 to the power of negative 29 end exponent space text Nm end text
None of the above
1.23 cross times 10 to the power of negative 23 end exponent space text Nm end text
0 space text Nm end text
1.32 cross times 10 to the power of negative 33 end exponent space text Nm end text
q6.png
Calculate the magnitude of the torque
1.92 cross times 10 to the power of negative 29 end exponent space text Nm end text
None of the above
1.23 cross times 10 to the power of negative 23 end exponent space text Nm end text
0 space text Nm end text
1.32 cross times 10 to the power of negative 33 end exponent space text Nm end text
To calculate the magnitude of the torque, we need to use the formula τ = pEsinθ, where τ is the torque, p is the dipole moment, E is the magnitude of the electric field, and θ is the angle between the dipole moment and the electric field.
Given:
q1 = +e = +1.6 x 10^-19 C
q2 = -e = -1.6 x 10^-19 C
d = 0.12 m
E = 10^5 N/C
The dipole moment, p, can be calculated as the product of one of the charges and the separation distance:
p = q1d = (1.6 x 10^-19 C)(0.12 m) = 1.92 x 10^-20 C·m
The angle between the dipole moment and electric field is 0 degrees, as they are parallel in this case.
Now, let's calculate the torque:
τ = pEsinθ = (1.92 x 10^-20 C·m)(10^5 N/C)(sin 0°) = 0 N·m
Therefore, the magnitude of the torque is 0 Nm.
Given:
q1 = +e = +1.6 x 10^-19 C
q2 = -e = -1.6 x 10^-19 C
d = 0.12 m
E = 10^5 N/C
The dipole moment, p, can be calculated as the product of one of the charges and the separation distance:
p = q1d = (1.6 x 10^-19 C)(0.12 m) = 1.92 x 10^-20 C·m
The angle between the dipole moment and electric field is 0 degrees, as they are parallel in this case.
Now, let's calculate the torque:
τ = pEsinθ = (1.92 x 10^-20 C·m)(10^5 N/C)(sin 0°) = 0 N·m
Therefore, the magnitude of the torque is 0 Nm.
Which of the following is a vector?
None of the above
Electric field
Electric potential energy
Electric potential
Equipotential lines
None of the above
Electric field
Electric potential energy
Electric potential
Equipotential lines
The electric field is a vector.
1 answer