# Electric field strength at the midpoint between the two charges.

## Homework Statement

A + 7.3 nC point charge and a - 2.4 nC point charge are 3.5 cm apart. What is the electric field strength at the midpoint between the two charges?

## Homework Equations

E = 9*10^9(q/r^2)
q = charge
r = distance from point charge

## The Attempt at a Solution

Since the question asks for the field strength between the two charges, r would be 1.75 cm or .0175 m.
Therefore

E = E1+E2
E1=9*10^9(7.3*10^-9/.0175^2)
E1=214531
E2=9*10^9(-2.4*10^-9/.0175^2)
E2=-70530.6
E=144000 N/C

Except it says I'm wrong. Its multiple choice, and the other choices given are
1900 N/C
5700 N/C
2.8×10^5 N/C

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rl.bhat
Homework Helper
If you keep a positive test charge at the mid point, positive charge will repel it and negative charge will attract it. So E1 and E2 are in the same direction. Hence

E = E1+E2 = 214531 + 70530.6 = ...?

So, to make this work, would my E2 equation have to be E=9*10^9(q/-r^2)? I don't know what you mean when you say E1 and E2 are in the same direction.

rl.bhat
Homework Helper
So, to make this work, would my E2 equation have to be E=9*10^9(q/-r^2)? I don't know what you mean when you say E1 and E2 are in the same direction.
The magnitude of E2 = k*q2/r2 and the direction is towards -2.4 nC.

The magnitude of E1 = k*q1/r2 and the direction is towards -2.4 nC.

where r = d/2.