Published by:
CGP EDU Academic Team
Published on: September 12, 2026
Electric potential is given by
$V = 6x - 8xy^{2} - 8y + 6yz - 4z^{2}$
Then electric force acting on $20^{\circ} \mathrm{C}$ point charge placed on origin will be
Text Solution
Verified by ExpertsThe correct answer is:
D
$E_x = -\frac{dV}{dx} = -(6 - 8y^2),$
$E_{y} = -\frac{dV}{dy} = -(-16xy - 8y + 6z)$
$E_z = -\frac{dV}{dz} = -(6y - 8z)$
At origin x = y = z = 0 so, $E_x = -6, E_y = 8$ and $E_{z} = 0$
⇒ ⇒ $E = \sqrt{E_x^2 + E_y^2} = 10 \, \mathrm{N/C}$ .
Hence force $F = QE = 2 \times 10 = 20\,\mathrm{N}$
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