Question

Charge of 4uc,-4uc and-8uc are placed at A,B and C corners of a square of side 10cm. Calculate the field intensity at side D

Answer 1

Given :

Magnitude of charge placed at corner  A = 4 $\mu C$

Magnitude of charge placed at corner  B = -4 $\mu C$

Magnitude of charge placed at corner  C = -8 $\mu C$

Side length of the square = 10 cm

To Find :

Intensity of electric field at  D = ?

Solution :

Since side length is 10 cm , so the length AD and CD is 10 cm  ,  but BD is diagonal so length of BD is :

$BD = \sqrt{10^2 + 10^2 } =10 \sqrt2$ cm

Now the electric field at D due to charge at A is :

=$\frac{K\times 4}{10^2}$

= $\frac{4K}{100}$ N\C

Electric field at D due to charge at C is :

=$\frac{K\times 8}{10^2}$

= $\frac{8K}{100}$ N\C

Electric field at D due to charge at B is :

= $\frac{K \times 4}{(10\sqrt2)^2}$

=$\frac{4K}{200}$ N\C

Now electric field at D is resultant of electric field due to all the three charges .

Resultant electric field due to charges at C and A is :

=$\sqrt{(\frac{8K}{100})^2 + (\frac{4K}{100})^2}$

=$\frac{4\sqrt5K}{100}$ N\C

Now net electric field at D will be :

=$\sqrt{(\frac{4\sqrt5K}{100})^2+(\frac{4K}{200})^2}$

=$\sqrt{(\frac{4\sqrt5K}{100})^2+(\frac{2K}{100})^2}$

=$\frac{2\sqrt{21}K}{100}$

=$9.16 \times 10^{-2}$ N\C

Hence , net electric field intensity at D is $9.16 \times 10^{-2}$ N/C .