Four charges $1 0^{- 8}; - 2 \times 1 0^{- 8}; + 3 \times 1 0^{- 8}$ and $2 \times 1 0^{- 8}$ coulombs are placed at the four corners of a square of side $1 m$ . The potential at the centre of the square is

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Answer 1

We know V due to a charge Q $= \frac{k Q}{r}$

Potential at centre $= \frac{k ( 1 0 ^{- 8} )}{( \frac{1}{2} )} + \frac{k ( - 2 \times 1 0 ^{- 8} )}{( \frac{1}{2} )} + \frac{k ( 2 \times 1 0 ^{- 8} )}{( \frac{1}{2} )}$
$+ \frac{k ( 3 \times 1 0 ^{- 8} )}{( \frac{1}{2} )}$

$= k 2 (1 0^{- 8} + 3 \times 1 0^{- 8})$
$= 4 \times 1 0^{- 8} \times 9 \times 1 0^{9} \times 2$
$= 3602 v o l t s$

Answer 2

We know V due to a charge Q

= \frac{k Q}{r}

Potential at centre

= \frac{k ( 1 0 ^{- 8} )}{( \frac{1}{2} )} + \frac{k ( - 2 \times 1 0 ^{- 8} )}{( \frac{1}{2} )} + \frac{k ( 2 \times 1 0 ^{- 8} )}{( \frac{1}{2} )}

+ \frac{k ( 3 \times 1 0 ^{- 8} )}{( \frac{1}{2} )}

= k 2 (1 0^{- 8} + 3 \times 1 0^{- 8})

= 4 \times 1 0^{- 8} \times 9 \times 1 0^{9} \times 2

= 3602 v o l t s

Four charges $1 0^{- 8}; - 2 \times 1 0^{- 8}; + 3 \times 1 0^{- 8}$ and $2 \times 1 0^{- 8}$ coulombs are placed at the four corners of a square of side $1 m$ . The potential at the centre of the square is

$z e r o$

$360 v o l t$

$180 v o l t$

$3602$ volt

Solution

$3602$ volt

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