Question
Three point charges Q+10x10-12C, Q₂= -6 x 10-12C, and Q3+10x10-12C, are situated at the corners of an equilateral triangle of side 10 m in vacuum. The electric potential at the centre of the triangle is
Answers
GPT 3.5
To find the electric potential at the center of the triangle, we first need to calculate the electric potential due to each individual charge and then sum them up.
The electric potential at a distance r from a point charge Q is given by the formula:
V = k(Q/r)
Where:
V = electric potential
k = Coulomb's constant (8.99 x 10^9 N m^2/C^2)
Q = charge
r = distance
Let's calculate the electric potential due to each charge individually:
For charge Q1 (+10 x 10^-12C):
V1 = k(Q1/r) = (8.99 x 10^9)(10 x 10^-12)/(5) = 17.98 V
For charge Q2 (-6 x 10^-12C):
V2 = k(Q2/r) = (8.99 x 10^9)(-6 x 10^-12)/(5) = -10.79 V
For charge Q3 (+10 x 10^-12C):
V3 = k(Q3/r) = (8.99 x 10^9)(10 x 10^-12)/(5) = 17.98 V
Now, we need to find the total electric potential at the center of the triangle by summing up the individual potentials:
Total electric potential = V1 + V2 + V3 = 17.98 V - 10.79 V + 17.98 V = 25.17 V
Therefore, the electric potential at the center of the equilateral triangle is 25.17 V.
The electric potential at a distance r from a point charge Q is given by the formula:
V = k(Q/r)
Where:
V = electric potential
k = Coulomb's constant (8.99 x 10^9 N m^2/C^2)
Q = charge
r = distance
Let's calculate the electric potential due to each charge individually:
For charge Q1 (+10 x 10^-12C):
V1 = k(Q1/r) = (8.99 x 10^9)(10 x 10^-12)/(5) = 17.98 V
For charge Q2 (-6 x 10^-12C):
V2 = k(Q2/r) = (8.99 x 10^9)(-6 x 10^-12)/(5) = -10.79 V
For charge Q3 (+10 x 10^-12C):
V3 = k(Q3/r) = (8.99 x 10^9)(10 x 10^-12)/(5) = 17.98 V
Now, we need to find the total electric potential at the center of the triangle by summing up the individual potentials:
Total electric potential = V1 + V2 + V3 = 17.98 V - 10.79 V + 17.98 V = 25.17 V
Therefore, the electric potential at the center of the equilateral triangle is 25.17 V.
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