The electric potential in a certain region is expressed by v = 6x-8xy $^{2}$ -8y+6yz-4z $^{2}$ volts. The magnitude of the force acting on a charge of 2 C situated at the origin will be

2 N

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6 N

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8 N

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20 N

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Solution

The correct option is D 20 N
Given expression for electric potential,
v=6x - 8xy $^{2}$ - 8y + 6yz - 4z $^{2}$ volts
charge Q=2 C
To find, Force (F).
We know,
F=QE...(Formula)
Where Q is charge and E is electric field,
We will solve given expression.
Along X-axis
$= \frac{d v}{d x} = 6 - 8 y^{2} . . . (1)$
Along Y-axis
$= \frac{d v}{d y} = - 16 x y - 8 + 6 z . . . (2)$
Along Z-axis
$= \frac{d v}{d z} = 6 y - 8 z . . . (3)$
At origin, from (1), (2) and (3)
$(\frac{d v}{d x}, \frac{d v}{d y}, \frac{d v}{d z}) = (6, - 8, 0)$
Net electric field,
$E = \sqrt{6^{2} + (- 8)^{2} + (0)^{2}} = \sqrt{100}$
$∴ E = 10 N / C$
$∴$ F= 2 x 10
$∴$ F=20 N
Option (D) is correct

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The electric potential in a certain region is expressed by v = 6x-8xy2-8y+6yz-4z2 volts. The magnitude of the force acting on a charge of 2 C situated at the origin will be

The electric potential in a certain region is expressed by v = 6x-8xy $^{2}$ -8y+6yz-4z $^{2}$ volts. The magnitude of the force acting on a charge of 2 C situated at the origin will be