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Idea of Symmetry

Charge of +q,...

Question

Charge of +q, +2q, +q and -q are placed at the corners of a square ABCD, calculate the electric field at the intersection of the diagonals of the square when $q = 5 / 3 \times 10^{- 9}$ C and each side square of is 10 cm.

2.1 \times 10^{3} V m^{- 1}

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9.1 \times 10^{2} V m^{- 1}

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1.8 \times 10^{4} V m^{- 1}

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0.9 \times 10^{4} V m^{- 1}

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Solution

The correct option is D $0.9 \times 10^{4} V m^{- 1}$
$E = k \times \frac{(q - q + q + 2 q)}{r^{2}}$

$= k . \frac{3 q}{r^{2}}$

$E = \frac{9 \times 10^{9} [3 \times \frac{5}{3} \times 10^{- 9}]}{(0.1 / \sqrt{2})^{2}}$

$= 9000 N / C$

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Similar questions

Four point charges $q_{A} = 2 μ C, q_{B} = - 5 μ C, q_{C} = 2 μ C, a n d q_{D} = - 5 μ C$ are located at the corners of a square ABCD of side 10cm. What is the force on a charge of $1 μ C$ placed at the centre of the square?

Q. Four charge

+ q, + q, - q

and

- q

are placed respectively at the corners

A, B, C

and

D

of a square of side

(a)

, arranged in the given order. Calculate the intensity of electric field at the center of the square where Q is a charge.

Q. Charges of

- 2 \times 10^{- 10} C, 2 \times 10^{- 10}

and

4 \times 10^{- 10} C

are placed at the corners A, B and D of a square ABCD of side 2m. Calculate the resultatnt field at the centre of the square.

Q. Four point charges

q_{A} = 2 μ C

q_{B} = 5 μ C

q_{C} = 2 μ C

q_{D} = 5 μ C

are located at the four corners of a square ABCD of side 10 cm. The force on a charge of

1 μ C

placed at the centre of the square is

Q. Point charges

+ q, + q, - q

and

- q

are placed at the corners A, B, C and D respectively of a square. O is the point of intersection of the diagonals AC and BD. The resultant electric field intensity at the point O.

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Charge of +q, +2q, +q and -q are placed at the corners of a square ABCD, calculate the electric field at the intersection of the diagonals of the square when q=5/3×10−9 C and each side square of is 10 cm.

The correct option is D 0.9×104Vm−1E=k×(q–q+q+2q)r2 =k.3qr2 E=9×109[3×53×10−9](0.1/√2)2 =9000N/C

Charge of +q, +2q, +q and -q are placed at the corners of a square ABCD, calculate the electric field at the intersection of the diagonals of the square when $q = 5 / 3 \times 10^{- 9}$ C and each side square of is 10 cm.

The correct option is D $0.9 \times 10^{4} V m^{- 1}$
$E = k \times \frac{(q - q + q + 2 q)}{r^{2}}$

$= k . \frac{3 q}{r^{2}}$

$E = \frac{9 \times 10^{9} [3 \times \frac{5}{3} \times 10^{- 9}]}{(0.1 / \sqrt{2})^{2}}$

$= 9000 N / C$