Question

Magnitude of point charge due to which electric field 30cm away has magnitude 2nc^-1 will be

Answer 1

hi,

electrric field due to apoint charge at a distance x from it

                =  1/4¶ε q/ x^2

              2  = 8.9 *10^9  *q  /  (.30)^2 

               q = 2* .09 / 8.9   * 10^-9

                  =  2.02 * 10^-11   C

Answer 2

Answer:

Charge, q = 2 × 10⁻¹¹ C

Explanation:

It is given that,

Electric field, E = 2 N/C

Distance, d = 30 cm = 0.3 m

Electric field is given by :

$E=\dfrac{kq}{d^2}$

k is the electrostatic constant

q is the point charge

$q=\dfrac{Ed^2}{k}$

$q=\dfrac{2\times (0.3)^2}{9\times 10^9}$

$q=2\times 10^{-11}$

The magnitude of point charge is 2 × 10⁻¹¹ C.