Question

A charge of 50 mu C placed at a point in an electric field is acted upon by a force of 10-2N .What is the intensity of the electric field at the point? a) 100N/C b) 150N/C c) 175N/C d) 200N/C​

Answer 1

The correct option regarding the intensity of the electric field at the point is d) 200 N/C.

Given:

A charge of 50 mu C placed at a point in an electric field is acted upon by a force of 10-2N.

To Find:

The intensity of the electric field at the point.

Solution:

To find the intensity of the electric field at the point we will follow the following steps:

As we know,

The electric field is defined as the force acting on a charged particle because of the electric field of another charged particle placed at some distance.

$electric \: field = \frac{force}{charge}$

Electric field or intensity is the same thing.

So,

$electric \: intensity = \frac{force}{charge}$

According to the question:

Force =

${10}^{ - 2} N$

Charge q =

$50uC = 50 \times {10}^{ - 6} C$

$(1u \: = {10}^{ - 6} )$

Now,

Putting values in the above formula we get,

$electric \: intensity = \frac{force}{charge} = \frac{ {10}^{ - 2} }{50 \times {10}^{ - 6} }$

$electric \: intensity = \frac{ {10}^{ - 2 + 6} }{50} = \frac{ {10}^{4} }{50} = \frac{10000}{50} = 200N {C}^{ - 1}$

Henceforth, the correct option regarding the intensity of the electric field at the point is d) 200 N/C.

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