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4 years ago

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Calculate the force on the test charge by the +6μC charge. Hint: The force in the positive x-direction is positive and the neg

ative x direction is negative.

1 answer:

DIA [1.3K]4 years ago

6 0

Answer:

The force on the test charge is 114.75 N.

Explanation:

Given that,

Test charge $q=6\ \mu C$

Suppose the test charge is 3.4 μC and the distance between the charge and test charge is 4.0 cm.

We need to calculate the force on the test charge

Using formula of electric force

$F=\dfrac{kqq_{1}}{r^2}$

Where, q = test charge

q₁ =charge

r = distance

Put the value into the formula

$F=\dfrac{9\times10^{9}\times6\times10^{-6}\times3.4\times10^{-6}}{(4.0\times10^{-2})^2}$

$F=114.75\ N$

Hence, The force on the test charge is 114.75 N.

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<u>Given</u>:

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