Question

Find the magnitude of the electric field experienced by a charge 15nanocoulomb which is kept 4cm away from a source charge within the electric field of the source charge​

Answer 1

Answer:

Magnitude of electric field = 8.4375 * 10⁴ N/C

Explanation:

Given that,

Charge = q = 15*10⁻⁹ C

Distance from charge = r = 4cm = 4*10⁻² m

We know that,

Electric field = E

$\boxed{E = \frac{kq}{r^{2} } }$  where, k = 9*10⁹

Putting the values in we get,

E = 9*10⁹*15*10⁻⁹ / (4*10⁻²)²

E = 135 / 16*10⁻⁴

E = 1350000 / 16

E = 84375 N/C

E = 8.4375 * 10⁴ N/C