SATHEE: Electrostatics Question 331

Electrostatics Question 331

Question: Charges of $+ \frac{10}{3} \times 10^{- 9} C$ are placed at each of the four corners of a square of side $8 c m$ . The potential at the intersection of the diagonals is [BIT 1993]

Options:

A) $150 \sqrt{2} v o l t$

B) $1500 \sqrt{2} v o l t$

C) $900 \sqrt{2} v o l t$

D) $900 v o l t$

Show Answer

Answer:

Correct Answer: B

Solution:

Potential at the centre O,

$V = 4 \times \frac{1}{4 π ε_{0}} . \frac{Q}{a / \sqrt{2}}$ where $Q = \frac{10}{3} \times 10^{- 9} C$

and $a = 8 c m = 8 \times 10^{- 2} m$

So $V = 5 \times 9 \times 10^{9} \times \frac{\frac{10}{3} \times 10^{- 9}}{\frac{8 \times 10^{- 2}}{\sqrt{2}}}$

$= 1500 \sqrt{2} v o l t$

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Electrostatics Question 331

Question: Charges of +103×10−9C are placed at each of the four corners of a square of side 8cm . The potential at the intersection of the diagonals is [BIT 1993]

Options:

Answer:

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Question: Charges of $+ \frac{10}{3} \times 10^{- 9} C$ are placed at each of the four corners of a square of side $8 c m$ . The potential at the intersection of the diagonals is [BIT 1993]