Question

The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the charge on the object?

Answer 1

Answer:

q = 4.5 nC

Explanation:

given,

electric field of small charged object, E = 180000 N/C

distance between them, r = 1.5 cm = 0.015 m

using equation of electric field

$E = \dfrac{kq}{r^2}$

k = 9 x 10⁹ N.m²/C²

q is the charge of the object

$q= \dfrac{Er^2}{k}$

now,

$q= \dfrac{180000\times 0.015^2}{9\times 10^9}$

      q = 4.5 x 10⁻⁹ C

      q = 4.5 nC

the charge on the object is equal to 4.5 nC

Answer 2

Answer:

The charge on the object is $4.5\times10^{-9}\ C$

Explanation:

Given that,

Electric field = 180000 N/C

Distance = 1.5 cm

We need to calculate the charge on the object

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Where, q = charge

r = distance

Put the value into the formula

$180000=\dfrac{9\times10^{9}\times q}{(1.5\times10^{-2})^2}$

$q=\dfrac{180000\times(1.5\times10^{-2})^2}{9\times10^{9}}$

$q=4.5\times10^{-9}\ C$

Hence, The charge on the object is $4.5\times10^{-9}\ C$