Question

25.
The ratio of electric field intensity at distance 5 cm to that at 10 cm from a point charge 5Q in air is
sous
A) 2:1
B)4:1
C) 1:4
D) 1:2​

Answer 1

Answer:

electric field intensity=kq1q2/d^2

Ratio= k5Qq2/(5) ^2/k5Qq2/(10) ^2

=k5Qq2/25/k5Qq2/100

=100/25=4/1

i.e.The ratio of electric field intensity is 4:1.

Answer 2

The answer is 1:4........