Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Two infinite sheets having charge densities σ1 and σ2 are placed in two perpendicular planes whose two-dimensional view is shown in Fig. The charges are distributed uniformly on the sheets in electrostatic equilibrium condition. Four points are marked I, II, III, and IV. The electric field intensities at these points are E→1,E→2,E→3 and E→4 respectively. The correct expression for the electric field intensities is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

a

|E→1| = |E→2|=σ12+σ222ε0≠|E→4|

b

|E→2| = |E→4|=σ12+σ222ε0

c

|E→1| = |E→2| = |E→3|= |E→4| =σ12+σ222ε0

d

none of these

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Using the principle of superposition E due to infinite plane sheet having surface charge density con its one face is σ2ε0.E→1=σ12ε0i+σ22ε0j, E→2=−σ12ε0i+σ22ε0j, E→3=−σ12ε0i−σ22ε0j, E→4=σ12ε0i−σ22ε0j
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon