Question

find the electric intensity at a point with a charge of 5 into 10 ki power minus 4 coulomb experience a force of 2.25 Newton

Answer 1

Heya!

E = F/q

F = 2.25N

$q = 5 \times {10}^{ - 4}$
E = 0.45 × 10^-4

Thanks!

Answer 2

Answer:

The electric field intensity at a point is $45\times10^{2}\ N/C$

Explanation:

Given that,

Charge $q =5\times10^{-4}\ C$

Force F = 2.25 N

The electric field intensity is the force per unit charge.

Formula of the electric intensity is defined as:

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

$E=\dfrac{2.25}{5\times10^{-4}}$

$E = 4500\ N/C$

$E = 45\times10^{2}\ N/C$

Hence, The electric field intensity at a point is $45\times10^{2}\ N/C$