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Q27PE

Expert-verifiedFound in: Page 664

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 ${\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{\times}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{5}}{\mathbf{}}{\mathit{N}}$ upward force on a ${\mathbf{-}}{\mathbf{1}}{\mathbf{.}}{\mathbf{75}}{\mathbf{}}{\mathit{\mu}}{\mathit{C}}{\mathit{a}}$ charge?**

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

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

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

From the definition of the electric field,

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

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

Substituting all known values,

$E=\frac{2.00\times {10}^{-5}N}{-1.75\mu C}\phantom{\rule{0ex}{0ex}}=\frac{2.00\times {10}^{-5}N}{\left(-1.75\mu C\right)\times \left(\frac{{10}^{-6}C}{1\mu C}\right)}\phantom{\rule{0ex}{0ex}}=11.43N/C$

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.

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