Question

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An electric field of magnitude 5.25 x 105 N/C points due east at a certain location. Find the magnitude and direction of the force on a-7.15 uc charge at this location. magnitude N direction --Select- Need Help? Read It

Answer 1

The electric field is related to the force exerted on the charge by the eq.

$ElectricField =\frac{Force}{charge}$

thus,

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

$\Rightarrow \: \: \: \vec{F}=q\vec{E}$

$\Rightarrow \: \: \: \vec{F}=(-7.15\times 10^{-6} \, C)(5.25\times 10^{5}N/C\, \hat{a}_{east})$

or

${\color{Blue} \vec{F}=3.75\, N \, \hat{a}_{west}}$

The magnitude of the force is 3.75 N and points due to west.