Solveeit Logo

Question

Physics Question on Electric charges and fields

Let E1(r),E2(r)E _{1}( r ), E _{2}( r ) and E3(r)E _{3}( r ) be the respective electric fields at a distance rr from a point charge Q,anQ , an infinitely long wire with constant linear charge density λ\lambda, and an infinite plane with uniform surface charge density σ\sigma. If E1(r0)=E2(r0)=E3(r0)E_{1}\left(r_{0}\right)=E_{2}\left(r_{0}\right)=E_{3}\left(r_{0}\right) at a given distance r0r_{0}, then

A

Q=4σπr02Q=4\sigma \pi r^2_0

B

r0=λ2πσr _{0}=\frac{\lambda}{2 \pi \sigma}

C

E1(r0/2)=2E2(r0/2)E _{1}\left( r _{0} / 2\right)=2 E _{2}\left( r _{0} / 2\right)

D

E2(r0/2)=4E3(r0/2)E _{2}\left( r _{0} / 2\right)=4 E _{3}\left( r _{0} / 2\right)

Answer

E1(r0/2)=2E2(r0/2)E _{1}\left( r _{0} / 2\right)=2 E _{2}\left( r _{0} / 2\right)

Explanation

Solution

Q4πε0r02=λ2πε0I0=σ2ε0\frac{ Q }{4 \pi \varepsilon_{0} r _{0}^{2}}=\frac{\lambda}{2 \pi \varepsilon_{0} I _{0}}=\frac{\sigma}{2 \varepsilon_{0}}
E1(r02)=Qπε0I02,E _{1}\left(\frac{ r _{0}}{2}\right)=\frac{ Q }{\pi \varepsilon_{0} I _{0}^{2}},
E2(r02)=λπε0I0,E _{2}\left(\frac{ r _{0}}{2}\right)=\frac{\lambda}{\pi \varepsilon_{0} I _{0}},
E3(r02)=σ2ε0E _{3}\left(\frac{ r _{0}}{2}\right)=\frac{\sigma}{2 \varepsilon_{0}}
E1(r02)\therefore E _{1}\left(\frac{ r _{0}}{2}\right)
=2E2(r02)=2 E _{2}\left(\frac{ r _{0}}{2}\right)