Find the electric potential at point A

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The two charges in the figure below are separated by d = 4.00 cm. (Let q1 = -12 nC and q2 = 27.5 nC.)

(a) Find the electric potential at point A.

(b) Find the electric potential at point B, which is halfway between the charges.
asked Jan 20, 2012 in Physics by awesome ~Expert~ (1,479 points)

1 Answer

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Best answer
To find the electric potential at A you would need to add up the electric potentials of the two point charges affecting A, namely q1 and q2.

The formula for electric potential of point charge is

V = (ke*q)/r

Now you must apply this equation to q1 and q2:


V1 = (ke*q1)/r   and  V2 = (ke*q2)/r


plug in the values which are:

where ke = 8.988*10^9 N*m^2/C^2 (found in textbook)

q1 = -12*10^-9  C

q2 = 27.5*10^-9 C

r = 4*10^-2 m

(*note i have already converted all units to SI units)


V1 = (8.988*10^9 N*m^2/C^2  *  -12*10^-9  C) / 4*10^-2 m   = -2696.4 V


V2 = (8.988*10^9 N*m^2/C^2  *  27.5*10^-9 C) / 4*10^-2 m  = 6179.26 V


Now add up V1 and V2 to get the total electric potential at point A.

V1 + V2 = (-2696.4 V) + (6179.26 V) = 3482.9 V ~ 3.48 kV


Part B is easy to find now that we have solved part A.  Since point B is at a distance half of the original distance from both charges, you would essentially be dividing the answer by (1/2), which is the same is multiplying by 2,

and so the electric potential at B is equal to 2*3.48 kV = 6.96 kV
answered Jan 26, 2012 by pokemonmaster ~Expert~ (3,856 points)
selected Jan 26, 2012 by awesome