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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# What is the magnitude and direction of the force exerted on a ${\mathbf{3}}{\mathbf{.}}{\mathbf{50}}{\mathbf{}}{\mathbit{\mu }}{\mathbit{C}}$ charge by a 250 N/C electric field that points due east?

The magnitude of the force is $8.75×{10}^{-4}N$ and it is directed towards East.

See the step by step solution

## Step 1: Electric field

A space around the charge in which any other test charge experiences some force is known as electric field. The expression for the strength of the electric field is,

$E=\frac{F}{q}----------\left(1.1\right)$

Here, F is the force experienced by the test charge, q and is the test charge.

## Step 2: Magnitude and direction of force

The magnitude of the force experienced by the test charge can be calculated using equation (1.1).

Rearranging equation (1.1) in order to get an expression for the force experienced by the test charge.

F = qE

Here, q is the charge on test charge $\left(q=3.50\mu C\right)$, and E is the strength of electric field (250 N/C).

Substituting all known values,

$F=\left(3.50\mu C\right)×\left(250N/C\right)\phantom{\rule{0ex}{0ex}}=\left(3.50\mu C\right)×\left(\frac{{10}^{-6}C}{1\mu C}\right)×\left(250N/C\right)\phantom{\rule{0ex}{0ex}}=8.75×{10}^{-4}N$

Hence, the magnitude of the force is $8.75×{10}^{-4}N$.

Since, the direction of the force and the electric field are same. Hence, the force is directed towards East.