A +35 "mu"C point charge is placed 32 cm from an identical +35 "mu"C charge. What is the electric potential at the point midway between the two charges?

2 Answers

crmhaske's profile pic

crmhaske | College Teacher | (Level 3) Associate Educator

Posted on

If the two charges are 32 cm from each other, than the distance of the mid point to each charge is r = 32/2 = 16 cm.

The equation for voltage (V) with respect to charge (Q) is

`V=(kQ)/r` , where k is Columb's constant`k=8.987552x10^9 (Nm^2)/C^2`

The electric potential at any point P is simply the summation of electric potentials at point P, therefore:

a) At the mid point:


V = 3.93e6 V = 3.93 MV

b) 12 cm closer to the second charge:

r1 = 16+12=28; r2=16-12=4


V = 9e6 V = 9 MV


Therfore, the work required is 2.5 J

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to use the formula of electric potential such that:

`V = 1/(4*pi*epsilon_0)*(q_1*q_2)/(r/2)`

Since the problem provides `q_1 = q_2 = 35muC` and` r = 32 cm` yields:

`V = 1/(4*pi*epsilon_0)*(35^2)/(32/2)`

`V = 1/(4*pi*epsilon_0)*(35^2)/(32/2)`

`V = 19.14*8.89 => V = 170.160 V`

Hence, evaluating the electric potential at the midpoint between the charges, yields `V = 170.160 V.`