1
Question
Two point charges q and -q are separated by a distance 2l, Find the flux of electric field strength vector across the circle of radius R placed with its centre coinciding with the midpoint of line joining the two charges in the perpendicular plane.
Open in App
Solution
E=kqx2+y2+l2x^i+y^j−l^k√x2+y2+z2−kqx2+y2+l2x^i+y^j+l^z√x2+y2+z2
=−24πϵ0ql^k(x2+y2+z2)3/2
ϕ=∫→E.d→A=∫Edxdy^k
=−12πϵ0ql∫dxdy(x2+y2+z2)3/2
By doing change of coordinates,
E=−12πϵ0ql∫rdθdr(r2+l2)3/2
=−12πϵ0ql2π∫R0rdr(r2+l2)3/2
By doing change of coordinates, u=r2+l2
E=−qlϵ0∫du2u3/2
After solving this integral finally,
E=qϵ0(1−1√1+(R/l)2)
=−24πϵ0ql^k(x2+y2+z2)3/2
ϕ=∫→E.d→A=∫Edxdy^k
=−12πϵ0ql∫dxdy(x2+y2+z2)3/2
By doing change of coordinates,
E=−12πϵ0ql∫rdθdr(r2+l2)3/2
=−12πϵ0ql2π∫R0rdr(r2+l2)3/2
By doing change of coordinates, u=r2+l2
E=−qlϵ0∫du2u3/2
After solving this integral finally,
E=qϵ0(1−1√1+(R/l)2)
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
