Question

A force of 4.5 acts on a charge of 7.5×10-4C. calculate the intensity of electric field at that point

Answer 1

Answer:

hey,here u go

Explanation:

here you go

mark it as brainliest if u think this answer deserves dude

Answer 2

Given: Force is 4.5 N.

It is acting on a charge of $\[7.5\times {{10}^{-4}}\]$C.

To find: The intensity of the electric field at the given point.

Solution:

Electrostatic force of a body is directly proportional to its charge. This means, if force is increased, the charge will also be increased.  

The intensity of an electric field at a particular point is defined as the force which is undergone by a charge that is placed at the point.

It has a SI unit of Newton/Coulombs.

Therefore, it can be said that:

Electric field = $\frac{Force}{Charge}$

$& \Rightarrow E=\frac{F}{q} \\ & \Rightarrow E=\left( \frac{4.5}{7.5\times {{10}^{-4}}} \right)N/C \\ & \Rightarrow E=\left( \frac{0.6}{{{10}^{-4}}} \right)N/C \\ & \Rightarrow E=\left( 0.6\times {{10}^{4}} \right)N/C \\ & \Rightarrow E=\left( 6\times {{10}^{3}} \right)N/C$

Hence, the intensity of the electric field at that point is $( 6\times {{10}^{3}} )$ N/C.