Question

two charges each equal to 2 microcoulomb are 0.5m apart . if both of them exist inside vacuum then the force between them is

Answer 1

two equal charges

q=2microcoulomb =2*10^-6C

r=0.5m

F=kqq/r^2

• putting the values
• k=9*10^9 (for vacuum)

F=9*10^-9*4*10^-12

0.5 * 0.5

your finally answer is .....

F=1.4 * 10^-19

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Answer 2

Answer:

The electrostatic force between the 2 given charges is 1.44 × 10⁻⁵ N.

Explanation:

Given - two charges, 2 μC each
             distance between the two charges - 0.5 m
             medium of existence - vacuum

To find - the force between the 2 charges

Formula - $F = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}$

Solution - like we have studied, the force between 2 electric charges is said to be directly proportional to the product of the magnitude of charges, and inversely proportional to the square of the distance between them.

$F \propto q_1q_2\\F \propto \frac{1}{r^2}$

so, we use the constant $\frac{1}{4\pi \epsilon_0}$, also called the proportionality constant of electrostatic forces, and get the aforementioned formula.

Now, we calculate the force between the 2 charges -

$F = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r^2}\\F = (9 \times 10^9) \frac{2 \times 10^-^6 \times 2 \times 10^-^6}{0.5 \times 0.5}\\F = (9 \times 10^9) \frac{4 \times 10^-^1^2}{25 \times 10^-^2}\\\\F = \frac{36 \times 10^-^5}{25}\\F = 1.44 \times 10^-^5\; N$

Thus, the electrostatic force between these 2 charges is, as we calculated, 1.44 × 10⁻⁵ N.

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