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 | Add Yours

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=8.987552x10^9((35x10^-6)/0.16+(35x10^-6)/0.16)`

V = 3.93e6 V = 3.93 MV

b) 12 cm closer to the second charge:

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

`V=8.987552x10^9((35x10^-6)/0.28+(35x10^-6)/0.04)`

V = 9e6 V = 9 MV

`W=qDeltaV=(0.5x10^-6)(9x10^6-3.93x10^6)=2.535`

Therfore, the work required is 2.5 J

**Sources:**

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.`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes