Question

Q 6/180 The electric field due to a point charge at a
distance 1 m from it is 90 N/C. The magnitude of the
charge is
1)10 με
2)1 μC
3)10 nC
4)100 pc

Answer 1

For point charge

E=V/r

Given 

V=15J/C

E=30N/C

Distance from point charge, r=?

Point charge, q=?

r=V/E=15/30=0.5m

V=4πε0rq

q=4πε0rV

q=9×1091×0.5×15

q=0.83×10−9C

q=0.83nC

I hope this helps! :))

Answer 2

Given:

The distance of point charge, r = 1 m

The electric field, E = 90 N/C

To Find:

The magnitude of the  point charge.

Calculation:

- We know that electric field intensity is given as:

E = (1/4πε₀) × q / r²

- Putting all the known values in this equation, we get:

⇒ 90 = 9 × 10⁹ × q / (1)²

⇒ q = 90 / (9 × 10⁹)

⇒ q = 10 × 10⁻⁹

⇒ q = 10 nC

- So, the correct answer is (3) 10 nC.