Two charges of 6 - C and 24 - C are placed 12 cm apart- The intensity of the electric field is zero- at a pointA- 4 cm from the smaller charge and the point insideB- 4 cm from the smaller charge and the point outsideC- 4 cm from the bigger charge towards the smaller chargeD- 12 cm from both charges

The correct option is A 4 cm from the smaller charge and the point inside x=d/√(q_2/q_1)+1=12/√(24/6)+1=12/2+1=12/3=4 cm x = 4 cm from the smaller charge, poin ...

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Standard XII

Physics

Electric Field Due to a Line of Charge Not on Its Axis

Two charges o...

Question

Two charges of 6 $μ C$ and 24 $μ C$ are placed 12 cm apart. The intensity of the electric field is zero, at a point

4 cm from the smaller charge and the point inside

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4 cm from the smaller charge and the point outside

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12 cm from both charges

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4 cm from the bigger charge towards the smaller charge

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Solution

The correct option is A 4 cm from the smaller charge and the point inside
$x = \frac{d}{\sqrt{\frac{q_{2}}{q_{1}}} + 1} = \frac{12}{\sqrt{\frac{24}{6}} + 1} = \frac{12}{2 + 1} = \frac{12}{3} = 4 c m$
x = 4 cm from the smaller charge, point inside

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Two charges of 6 μC and 24 μC are placed 12 cm apart. The intensity of the electric field is zero, at a point

The correct option is A 4 cm from the smaller charge and the point inside x=d√q2q1+1=12√246+1=122+1=123=4 cm x = 4 cm from the smaller charge, point inside

Two charges of 6 $μ C$ and 24 $μ C$ are placed 12 cm apart. The intensity of the electric field is zero, at a point

The correct option is A 4 cm from the smaller charge and the point inside
$x = \frac{d}{\sqrt{\frac{q_{2}}{q_{1}}} + 1} = \frac{12}{\sqrt{\frac{24}{6}} + 1} = \frac{12}{2 + 1} = \frac{12}{3} = 4 c m$
x = 4 cm from the smaller charge, point inside