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Expert-verified Found in: Page 664 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # What is the magnitude and direction of an electric field that exerts a ${\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{5}}{\mathbf{}}{\mathbit{N}}$ upward force on a ${\mathbf{-}}{\mathbf{1}}{\mathbf{.}}{\mathbf{75}}{\mathbf{}}{\mathbit{\mu }}{\mathbit{C}}{\mathbit{a}}$ charge?

The magnitude of the electric field is $11.43N/C$ , and the direction is downward.

See the step by step solution

## Step 1: Given Data

Electric field E exerts a force $F\left(=2.00×{10}^{-5}N\right)$on a charge $\left(=-1.75\mu C\right)$.

## Step 2: Electric field

The force per unit positive test charge is known as the intensity of the electric field.

## Step 3: Magnitude and direction of electric field

From the definition of the electric field,

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

Here, F is the force $\left(F=2.00×{10}^{-5}N\right)$ , and q is the test charge $\left(q=1.75\mu C\right)$ .

Substituting all known values,

Hence, the magnitude of the electric field is 11.43 N/C.

The negative sign of the electric field indicates that the direction of electric field is in the opposite direction of force. Since, the force is directed upward. Hence, the electric field is directed downward. ### Want to see more solutions like these? 