# 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?

crmhaske | Certified Educator

calendarEducator since 2010

starTop subjects are Math, Science, and Literature

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

check Approved by eNotes Editorial

## Related Questions

sciencesolve | Student

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.`