Question

1. Two charges

3× 10^-5
and

5×10
^5
are placed at a distance 10 cm from each other. Find the

value of electrostatic force acting between them.


Answer it fast and explain the calculation part​

Answer 1

Explanation:

f=kq1q2/d^2

where the k is the constant and its value is 9×10^9

=9×10^9×3×10^-5×5×10^5/100×10^-4

=135×10^11

if ther is any mistake in calculations then sorry but you should remember the concept

Answer 2

Question :-:

Given above ↑

Answer :-:

$\to \: \boxed{\sf F = 13.5 \times {10}^{11} } \:$

To find :-:

→ Electrostatic force (F)

Formula used :-

$\to \boxed{ \sf F = k \frac{ \: q_1 \: q_2\: }{ {r}^{2} } } \\ \\ \sf k \: is \: constant \: of \: value \: 9 \times {10}^{9}$

Explanation :-

Given that

→ distance between these forces ( r) is 10 cm = 0.1 m

apply the given formula

$\to \sf F = 9 \times {10}^{9} \times \frac{3 \times {10}^{ - 5} \times 5 \times {10}^{4} }{0.1 \times 0.1} \\ \\ \to \boxed{\sf F = 135 \times {10}^{10} } \\ \\ \to \: \boxed{\sf F = 13.5 \times {10}^{11} } \:$