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Q28PE

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 the force exerted on a ${\mathbf{3}}{\mathbf{.}}{\mathbf{50}}{\mathbf{}}{\mathit{\mu}}{\mathit{C}}$ charge by a 250 N/C electric field that points due east?**

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

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,

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

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

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,

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

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

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

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