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Electrostatics Part 2

Section: Physics

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Question : 100 of 100

Marks: +1, -0

Find the electric field at point

P

(as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length

L

carrying a charge

Q

. The distance of the point

P

from the centre of the rod is

a=\frac{\sqrt{3}}{2}L

[26 Feb 2021 Shift 1]

$\frac{\sqrt{3}Q}{4\pi\varepsilon_0 L^2}$
$\frac{Q}{3\pi\varepsilon_0 L^2}$
$\frac{Q}{2\sqrt{3}\pi\varepsilon_0 L^2}$
$\frac{Q}{4\pi\varepsilon_0 L^2}$

Solution:

Given, length of conductor

=L

Charge on conductor

=Q

According to figure,

OP=a=\frac{\sqrt{3}}{2}L, OQ=\frac{L}{2}

Let

PQ=r=\sqrt{OP^{2}+OQ^{2}}

\Rightarrow PQ=\sqrt{\left(\frac{\sqrt{3}}{2}L\right)^2+\left(\frac{L}{2}\right)^2}=\sqrt{\frac{3}{4}L^2+\frac{L}{4}}=L

and

E

be the electric field at point

P

.

Since,

E

(due to finite wire)

=\frac{k\lambda}{a}(\sin\varphi_1+\sin\varphi_2)

. . . (i)

where,

k=

Coulomb's constant

=\frac{1}{4\pi\varepsilon_0}

\lambda=

linear charge density

=\frac{Q}{L}

and

\sin\varphi=\sin\varphi_2=\frac{\frac{L}{2}}{L}=\frac{1}{2}

Substituting the above value in Eq. (i), we get

E=\frac{k\lambda}{a}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{L\cdot\frac{\sqrt{3}L}{2}}=\frac{1}{2\sqrt{3}\pi\varepsilon_0 L^2}

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