Question

The electric field at a point due to a point charge 25cm away is 10 n/c . What is the magnitude of charge

Answer 1

Please refer to the attachment for the solution.......

Hope u get that right.....

Answer 2

Answer:

The magnitude of the charge is 6.94×10⁻¹²C.

Explanation:

The electric field due to a point charge is given as,

$E=\frac{kQ}{r^2}$          (1)

Where,

E=electric field due to a point charge

k=coulomb's constant=9×10⁹kg.m³.s⁻².C⁻²

Q=magnitude of charge on the point charge

r=distance at which electric field is to be found

From the question we have,

E=10N/C

r=25cm=25×10⁻²m         (1cm=10⁻²m)

By substituting the required values in equation (1) we get;

$10=\frac{9\times 10^9\times Q}{(25\times 10^-^2)^2}$

$Q=\frac{10\times (25\times 10^-^2)^2}{9\times 10^9}$

$Q=\frac{10\times 625\times 10^-^4}{9\times 10^9}$

$Q=6.94\times 10^-^1^2C$

Hence, the magnitude of the charge is 6.94×10⁻¹²C.