Question

Two equal charges of +2x10^-16 C are kept 20cm apart in air. At point midway b/w them find (i) electric field intensity (ii) force acting on a charge of +2x10^-16 C.

Answer 1

Answer:

Part i)

Electric field intensity at mid point is ZERO

Part ii)

Force on each charge due to other charge is

$F = 9 \times 10^{-21} N$

Explanation:

Part i)

As we know that two charges are equal in magnitude as well as in sign

So here the electric field at mid point of two charges will cancel each other

so net electric field at mid point must be ZERO

Part ii)

Force between two charges is given by coulomb's law

it is given as

$F = \frac{kq_1q_2}{r^2}$

so we have

$F = \frac{(9\times 10^9)(2\times 10^{-16})(2\times 10^{-16})}{0.20^2}$

$F = 9 \times 10^{-21} N$

#Learn

Topic : Electrostatic force

https://brainly.in/question/15730060

Answer 2

Answer:

let the charges are q1,q2

so q1=q2=2*10^(-16)C

distance,d=20cm so radius,r=d/2=5cm=5*10^(-2)m

force F=(10^(-9)*2*10^(-16)*2*10^(-16))/(9*25*10^(-4))

            =1.77*10^(-39)N

Therefore electric field intensity, E=F/q

                                                         =8.88*10^(-24)volts