Infinite Sheet
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Q.
A non-conducting square sheet of side 10 m is charged with a uniform surface charge density, σ=−60μCm2 . Find the magnitude and orientation of electric field vector due to the sheet at a point which is d = 0.02 mm away from the midpoint of the sheet.
3.39 x 106 N/C, away from the sheet
6.78 x 106 N/C, towards the sheet
3.39 x 106 N/C, towards the sheet
6.78 x 106 N/C, away from the sheet
Q. Figure shows a large conducting ceiling having uniform charge density σ below which a charge particle of charge q0 and mass m is hung from point O, through a small string of length L. Calculate the minimum horizontal velocity v required for the string to become horizontal.


Q. Figure shows a large conducting ceiling having uniform charge density σ below which a charge particle of charge q0 and mass m is hung from point O, through a small string of length L. Calculate the minimum horizontal velocity v required for the string to become horizontal.


- v=√2mgl+2q0×σ2ϵ0×Lm
- v=√2mgl+2q0×σ2ϵ0×Lm
- v=√2mgl+2q0×σ2ϵ0×Lm
- v=√2mgl2+2q0×σ2ϵ0×L2mL2
Q. Two thin infinite parallel sheets have uniform surface densities of charge +σ and −σ. Electric field in the space between the two sheets is:
- σ/ϵ0
- σ/2ϵ0
- zero
- 2σ/ϵ0
Q. Three infinitely long charge sheets are placed as shown in the figure. The electric field at point P is going to be


- 2σε0^k
- −2σε0^k
- 4σε0^k
- −4σε0^k
Q. The line AA' is on a changed infinite conducting plane which is perpendicular to the plane of the paper. The plane has a surface density of charge σ and B is a ball of mass m with a like charge of magnitude q. B is connected by a string from a point on the line AA'. The tangent of the angle (θ) formed between the line AA' and the string is:
- qσ2ϵ0mg
- qσ4πϵ0mg
- qσ2πϵ0mg
- qσϵ0mg
Q. Find the electric field at −0.5d<z<0.5d

- E=0
- E=−σϵ^k
- E=σϵ^k
- E=−σ2ϵ^k
Q.
Two thin infinite parallel sheets have uniform surface densities of charge +σ and −σ. The value of E in the space outside the sheets is:
- σ2ϵ0
- σϵ0
- 2σϵ0
- zero
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 sheet in electrostatic equilibrium condition. Four points are marked I, II, III, and IV. The electric field intensities at these points are →E1, →E2, →E3, and→E4, respectively. The correct expression for the electric field intensities is :


- |→E1|=|→E2|=√σ21+σ222ε0≠|→E4|
- |→E2|=|→E4|=√σ21+σ222ε0
- |→E1|=|→E2|=|→E3|=|→E4|=√σ21+σ222ε0
- none of these
Q. The electric field outside the plates of two oppositely charged plane sheets of charge density σ is :
- +σ2ε0
- σε0
- Zero
- −σ2ε0