Question

Force acting on 3.2x10³C plced at a point in a uniform electric field is 0.128 N. The intensity of electric field is N/C. [a] 0.4 [b] 400 [c] 4000 [d] 40​

Answer 1

• Charge=Q=3.2×10^3C
• Force=F=0.128N

We know

$\boxed{\sf Intensity\:of\:electric\:field=\dfrac{Force}{Charge}}$

$\\ \sf\longmapsto Intensity\:of\:electric \:field=\dfrac{0.128}{3.2\times 10^{3}}$

$\\ \sf\longmapsto Intensity\:of\:electric\:field=40N/C$