# Find the electric potential at point A

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.
kV

(b) Find the electric potential at point B, which is halfway between the charges.
kV
asked Jan 20, 2012 in Physics

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 ~Expert~ (3,856 points)
selected Jan 26, 2012 by awesome