SATHEE: Electrostatics Question 774

Electrostatics Question 774

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

Options:

A) $(10 / 9) v o l t / μ m$ and in the +ve x direction

B) $(5 / 3) v o l t / μ m$ and in the -ve x direction

C) $(5 / 3) v o l t / μ m$ and in the +ve x direction

D) $(10 / 9) v o l t / μ m$ and in the -ve x direction

Show Answer

Answer:

Correct Answer: A

Solution:

[a] Here, $V (x) = \frac{20}{x^{2} - 4} v o l t$

We know that $E = - \frac{d V}{d x} = - \frac{d}{d x} (\frac{20}{x^{2} - 4})$

$or, E = + \frac{40 x}{{(x^{2} - 4)}^{2}}$

At $x = 4 μ m$ ,

$E = + \frac{40 \times 4}{{(4^{2} - 4)}^{2}} = + \frac{160}{144} = + \frac{10}{9} v o l t / μ m$ .

Positive sign indicates that $\vec{E}$ is in +ve x-direction.

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Electrostatics Question 774

Question: The potential at the point x (measured inμm ) due to some charges situated on the x-axis is given by V(x)=20(x2−4) volt. The electric field E at x=4μm

Options:

Answer:

Solution:

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