Question

hey mate can anyone answer this??​

Answer 1

$f = kq \div {r}^{2} \\ f = 9 \times {10}^{9} \times 5 \times {10}^{ - 6} \div {0.1}^{2} \\ f = 45 \times {10}^{3} \div {10}^{ - 2} \\ f = 4.5 \times {10}^{6}$

so,force experienced will be

2F×cos(Ѳ/2)

2×4.5×10^6×cos30°

=7.8×10^6 N

hope it helps!!

regards,

shreyas...

$\huge\mathcal\red{thank you}$