NEET 2023 — Physics solution

NEET2023Physics

An electric dipole is placed as shown in the figure.

The electric potential (in $1 0^{2}$ V) at point P due to the dipole is ( $ε_{0} =$ permittivity of free space and $\frac{1}{4 π ε _{0}} = K$ ):

A
$(\frac{5}{8}) q K$
B
$(\frac{8}{5}) q K$
C
$(\frac{8}{3}) q K$
D
$(\frac{3}{8}) q K$

Solution

Distance of P from $+ q$ : $5 - 3 = 2$ cm $= 2 \times 1 0^{- 2}$ m.

Distance of P from $- q$ : $5 + 3 = 8$ cm $= 8 \times 1 0^{- 2}$ m.

Potential at P:

$V_{P} = \frac{K q}{2 \times 1 0 ^{- 2}} - \frac{K q}{8 \times 1 0 ^{- 2}} = K q (\frac{1}{2} - \frac{1}{8}) \times 1 0^{2} = \frac{3}{8} q K \times 1 0^{2}$ V.

So the potential in units of $1 0^{2}$ V is $\frac{3}{8} q K$ .

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