JEE Main 2022 · 29 June · evening · Single correct · +4 / −1
If the electric potential at any point (x, y, z) m in space is given by V = 3x² volt. The electric field at the point (1, 0, 3) m will be :
- (a) 3 Vm⁻¹, directed along positive x-axis.
- (b) 3 Vm⁻¹, directed along negative x-axis.
- (c) 6 Vm⁻¹, directed along positive x-axis.
- (d) 6 Vm⁻¹, directed along negative x-axis.
Answer and solution
Answer (d) 6 Vm⁻¹, directed along negative x-axis.
Solution
Physics sir · 4 steps
The field is minus the slope of V: differentiate 3x² along x and do not forget that minus sign.
- Field from potential. The electric field is the negative gradient of the potential. Each component is minus the rate of change of V along that axis.
E⃗ = -((∂ V)/(∂ x) î + (∂ V)/(∂ y) ĵ + (∂ V)/(∂ z) k̂)
- Differentiate. V = 3x² has no y or z in it, so E_y = E_z = 0. Along x, (∂ V)/(∂ x) = 6x, so Eₓ = -6x.
- Put in the point. At (1, 0, 3), x = 1, so E⃗ = -6 î V m⁻¹. The z = 3 does not matter because V does not depend on z.
- So. The field has magnitude 6 V m⁻¹ and points along the negative x-axis, option (d).
The idea to keep. Always use E⃗ = -∇ V. The field points the way the potential falls fastest, so a rising potential means the field points backwards.
Answer from the NTA final answer key. Solution by PrepWiser.