Question

Two positive charges of 6.0 x 10^-6 C are separated by 0.50 m. What is the magnitude of the force between the charges?

Answer 1

Answer:

F = 0.648 \: {C}^{2} \div {m}^{2}

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Explanation:

Given:

Two positive charges having charge

$6 \times {10}^{ - 6} \: C$

say:

$q1 = q2 = q = 6 \times {10}^{ - 6} \: C$

These charges are separated by distance say

$d = 0.5 \: m$

To Find:

Force F between these charges.

Formula:

$F = \frac{1}{4\pi \epsilon} \frac{(q1)(q2)}{ {r}^{2} }$

Solution:

$F = \frac{1}{4\pi \epsilon} \frac{(q1)(q2)}{ {r}^{2} } \\ as \: \frac{1}{4\pi \epsilon} = 9 \times {10}^{ - 9} \\ F = \frac{9 \times {10}^{ - 9} \times {q}^{2} }{ {0.5}^{2} } \\ F = \frac{9 \times {10}^{ - 9} \times {(6 \times {10}^{ - 6}) }^{2} }{ {0.5}^{2} } \\ F = 0.648 \: {C}^{2} \div {m}^{2}$