Question

two point charges q1 and q2 of 2 into 10 to the power minus 8 coulomb and minus 2 into 10 to the power minus 8 coulomb respectively are placed 0.4 metre apart calculate the electric field at the centre of the line joining the two charges

Answer 1

Answer:

Explanation:

q1=2*10^-8C

q2=-2*10^-8C

E=kq/r^-2

E=9*10^9*2*10^-8C/16*10^-2

E=1125

Answer 2

Given that,

Charge 1, $q_1=2\times 10^{-8}\ C$

Charge 2, $q_2=-2\times 10^{-8}\ Cc$

Distance between charges, d = 0.4 m

To find,

The electric field at the centre of the line joining the two charges.

Solution,

Electric field due to charge 1 is :

$E_1=\dfrac{kq_1}{r^2}\\\\E_1=\dfrac{9\times 10^9\times 2\times 10^{-8}}{(0.4)^2}\\\\E_1=1125\ N$

Electric field due to charge 2 is :

$E_2=\dfrac{kq_2}{r^2}\\\\E_2=\dfrac{-9\times 10^9\times 2\times 10^{-8}}{(0.4)^2}\\\\E_2=-1125\ N$

So, the net electric field at the centre of the line joining the two charges is equal to :

$E_{net}=E_2-E_1\\\\E_{net}=0$

Learn more,

Electric field

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