Question

Point charges of + 6 nC, - 6 nC and + 16 nC are placed at the corners A, C and D of a square ABCD of side 0.4 m. Find the electric field strength at B.​

Answer 1

Given

Point charges at A=+ 6 N/C,

Point charges at C=- 6 N/C

Point charges at D=+ 16 N/C

To Find:  $E_{B} \ = ?$

Solution

Following are the attachment of diagram

We know that

$E_{A}\ = \frac{k \ \alpha _{A} }{q^{2} }$

$=\frac{\ 9\ *\ 10^{9}\ (6 * 10^{-9} ) }{0.4^{2} } \\E_{A}\ =\ 337.5\ N/ C$

Now we calculated the

$E_{C} \ = \frac{\ 9\ *\ 10^{9}\ (-6 * 10^{-9} ) }{0.4^{2} } \\=\ -337.5\ N/C$

Now calculated the

$E_{D}\ = \frac{\ 9\ *\ 10^{9}\ (16 * 10^{-9} ) }{0.4^{2} }\\ = 450 N/C$

The electric field strength at B. can be determined by the

$E_{B} \ = \sqrt{\((E_{D}\ Cos\ 45\ + E_{A}\ )\ +\ ( E_{D}\ Sin \ 45 \ +\ E_{C}\ ) ^{2} }$

On putting the value we get

$E_{B} =\ 655.98\ N/C$