The potential at a point x (measured in μm ) due to some charges situated on the x -axis is given by V(x) = 20/(x^2 - 4) Volts. The electric field E a - Physics MCQ AIEEE 2007 2026 | ExamTest.live

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The potential at a point x (measured in μ m) due to some charges situated on the x -axis is given by V (x) = 20/ (x^{2} - 4) Volts. The electric field E at x = 4 μ m is given by:

Solution & Step-by-step Explanation

The relationship between electric field and potential is E = - \frac{d V}{d x} . V (x) = 20 (x^{2} - 4)^{- 1} \frac{d V}{d x} = 20 \cdot (- 1) (x^{2} - 4)^{- 2} \cdot (2 x) = \frac{- 40 x}{( x ^{2} - 4 ) ^{2}} Therefore, E = - (\frac{- 40 x}{( x ^{2} - 4 ) ^{2}}) = \frac{40 x}{( x ^{2} - 4 ) ^{2}} .At x = 4 μ m : E = \frac{40 \times 4}{( 4 ^{2} - 4 ) ^{2}} = \frac{160}{( 16 - 4 ) ^{2}} = \frac{160}{1 2 ^{2}} = \frac{160}{144} E = \frac{10}{9} Volt/ μ m Since the value is positive, the field is in the positive x direction.

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The potential at a point x (measured in μ m) due to some charges situated on the x -axis is given by V (x) = 20/ (x^{2} - 4) Volts. The electric field E at x = 4 μ m is given by:

5/3 Volt/ μ m and in the - ve x direction

5/3 Volt/ μ m and in the + ve x direction

10/9 Volt/ μ m and in the - ve x direction

10/9 Volt/ μ m and in the + ve x direction

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