Question

A -5.00 nC point charge is at the origin, and a second +6.00 nC point charge is on the x-axis at x = 0.500 m. Find the NET electric field at x = -1.0 m?

Answer 1

The electric field at a point in space from a point charge is given by

$$E = k \frac {q}{r ^ {2}}$$

So,

$$E _ {1} = k \frac {q _ {1}}{r _ {1} ^ {2}} \quad E _ {2} = k \frac {q _ {2}}{r _ {2} ^ {2}}$$

Then

$$E = E _ {1} + E _ {2} = k \frac {q _ {1}}{r _ {1} ^ {2}} + k \frac {q _ {2}}{r _ {2} ^ {2}} = k \left(\frac {q _ {1}}{r _ {1} ^ {2}} + \frac {q _ {2}}{r _ {2} ^ {2}}\right)$$ $$E = 9 * 1 0 ^ {9} \left(\frac {- 5 * 1 0 ^ {9}}{1 ^ {2}} + \frac {6 * 1 0 ^ {9}}{1 . 5 ^ {2}}\right) = - 9 \left(\frac {N}{C}\right)$$

So, net field is $9\mathrm{N} / \mathrm{C}$ and directed along the positive side of x-axis