Question

two points charge q1=0.2C and q2=0.4C are placed 0.1m apart. calculate the electric field at :- 1) mid point between the charge. 2)a point on the line joining q1 and q2 such that it is 0.5m away from q2 and 0.15m away from q1. ​

Answer 1

Explanation:

<---r=0.05m-----><----r^2=0.05m------>

Aq^1--------<E^2---------------.o----------->E^1-------.Q^2.B

<------------0.1m------------------>

Electric field due to q1= -E^1=1/4PI E 0.2/0.05^2

=9*10^9 x 0.2/0.05^2 = 720 x 10^9 N/C

ELectric field due to q2-e^2 = 1/4PI E 0.2/0.05^2

=9*10^9 x 0.4/0.05^2 = 1440 x 10^9 N/C

Resultant Electric field at mid-point-E= e1+e2

Since the net electric field is acting in opposite direction we have E= 1440 x10^9-720x10^9 N=720 X 10^9 N/C

A ._______B.<-R^2=0.05M->P-->E1

<-----0.1m--------><------0.5m-----> -----> e^2

<-------r^2=0.15m--------------------->

Let point P be on the line joining the charges such that it is 0.05m away from q2 and 0.15 m away from q1.

Electric field due to q1=  9x10^9 x 0.2/0.15^2=80x10^9  N/C

Electric field due to q2= 9 X10^9 X 0.4/0.05^2 = 1440x10^9 n/c

Since, Electric field is acting in the same direction

80x 10^9 +1440 x10^9

= 1520x 10^9

=15.2 x 10^11 N/C

Answer 2

Explanation:

I hope it will help you:)