Question

Three equal charges of 10x10* C
respectively, each located at the comers
of a right triangle whose sides are 15 cm,
20cm and 25cm respectively. Find the
force exerted on the charge located at the
90° angle.​

Answer 1

Answer:

As you can see

Resultant will be

$- (f _{2} ) \hat i - f _ 1 \hat j$

where

$f _{1} = k \frac{10 \times 10}{ {10}^{2} } = k$

$f _{2} = k \frac{10 \times 10}{15 \times 15} = \frac{4k}{9}$

So magnitude of resultant

$\sqrt{ {k}^{2} + {( \frac{4k}{9}) }^{2} } = 9 \times {10}^{9} \sqrt{ \frac{97}{81} } \approx \: {10}^{10} n$

Where