Question

THE ELECTRIC FIELD DUE TO A POINT CHARGE AT A DISTANCE 6m FROM IT IS 630N/C. FIND THE MAGNITUDE OF CHARGE.

Answer 1

Here E = 630 N/C , r = 6m , To find : Q

We know that

E = k Q / r^2 where k = coulomb's Constant = 9 x 10 ^9

=> Q = E r^2 / k

=> Q = 630 x 6 × 6 / 9 x 10 ^9

= 70 x 36 / 10 ^9 ( On dividing 630 by 9 )

= 2520 x 10 ^ -9 ( here 1 / 10 ^9 = 10 ^ -9 )

= 2.52 x 10 ^3 x 10 ^ -9 ( here 420 = 4.2 x 10 ^2)

= 2.52 x 10 ^ 3 + ( -9)

= 2.52 x 10 ^ 6 C = 2.52 uC

Hope this help......

Answer 2

Answer:

Explanation:

Electric field=630N/C=F

r=6

k=9*10^9

(Electric fields are basically the lines of force)

By Columb's law

r²=36

F=Kq/r²

q=Fr²/k

q=630*36/9*10^9

q=2.52*10^-6

q=2.52micro coulomb