Question

The force exerted by an electric field on charge 10 micro c at a point is 16×10raised to -4 N. What is intensity of electric field at that point?

Answer 1

160

the force exerted by an electric field is equal to multiplication of charge and electri field intensity

F=qE

E=F/q

E=16*10^-4/10^-5

E=160

Answer 2

Given:

• Force on the charge = 16 × 10^(-4) N.
• Charge = 10 micro-C

To find:

• Electric field intensity at that point ?

Calculation:

The electric field intensity at any point (with respect to a reference) is equal to the force experienced by a unit positive charge.

• In mathematical terms , we can say:

$E = \dfrac{F}{q}$

$\implies E = \dfrac{16 \times {10}^{ - 4} }{10 \times {10}^{ - 6} }$

$\implies E = \dfrac{16 \times {10}^{ - 4} }{ {10}^{ - 5} }$

$\implies E = 16 \times {10}^{ - 4 + 5}$

$\implies E = 16 \times 10$

$\implies E = 16 0 \: N/C$

So, the field intensity at that point is 160 N/C.