Explanation
An electric field is non zero on the axis of hallow current carrying wire.
Magnitude of ϵ will be same while its direction get reverse and V remains unchanged.
We know that No. of lines of force is proportional to the magnitude of charge and electric lines of force can never intersect. So, option (2) and (3) is not possible.
Electric lines of force starts from positive charge and ends at negative charge. So, in the question A is positive charge and B is negative charge. Density of line at A is more than that of B.
The electric field intensity at a point in an electric field is equal to the negative potential in a given direction.
-dv/dx = ϵ
Negative sign shows that potential decreases in the direction of electric field.
Dipole energy, U = -pE cosθ
For U to be maximum
Cosθ = -1
→ θ = π
Axis of electric dipole is always directed from negative charge to the positive charge.
Gaussian surface is a symmetrical closed surface containing charge distribution at every point of which electric field has a single value.
The flux passing through the square of 1m placed in x – y plane inside the electric field is zero because by Gauss theorem we can say that closed circuit not formed.
Gauss’s theorem is not applicable in gravitation. So only this statement is correct.
PE of a dipole is calculated as
U = -p.E = -pEcosθ
Greater the orientation, value of U will be more positive.
Decreasing order of orientation
(IV) > (III) > (I) > (II)
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