Question

The electric intensity at a point 3 metre away from a charge of 50 coulomb kept in air is​

Answer 1

Answer:

Electric field are vector quantities in this case they are acting in opposite direction.

E due to A=

r2

kq

=

(0⋅3)

2

(9×10

9

)(50μC)

(

x

^

)

E due to B=

(0⋅3)

2

(9×10

9

)(100μC)

(−

x

^

)

∴Net E=

(0⋅3)

2

9×10

9

(50−100)×10

−6

x

^

E=5×10

6

(−

x

^

) V/m