SATHEE: Electrostatics 5 Question 18

Electrostatics 5 Question 18

20. A combination of capacitors is set-up as shown in the figure. The magnitude of the electric field, due to a point charge $Q$ (having a charge equal to the sum of the charges on the $4 μ F$ and $9 μ F$ capacitors), at a point distant $30 m$ from it, would equal to

(2016 Main)

(a) $240 N / C$

(b) $360 N / C$

(d) $480 N / C$

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Answer:

Correct Answer: 20. (c)

Solution:

$3 μ F$ and $9 μ F = 12 μ F$

$\begin{aligned} 4 μ F and 12 μ F & = \frac{4 \times 12}{4 + 12} = 3 μ F \\ Q & = C V = 3 \times 8 = 24 μ C (on 4 μ F and 3 μ F) \end{aligned}$

Now, this $24 μ Cdistributes$ in direct ratio of capacity between $3 μ F$ and $9 μ F$ . Therefore,

$\begin{aligned} Q_{9 μ F} & = 18 μ C \\ ∴ Q_{4 μ F} + Q_{9 μ F} = 24 + 18 & = 42 μ C = Q \\ E & = \frac{k Q}{R^{2}} = \frac{9 \times 10^{9} \times 42 \times 10^{- 6}}{30^{2}} \\ = 420 N / C \end{aligned}$

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Electrostatics 5 Question 18

20. A combination of capacitors is set-up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4μF and 9μF capacitors), at a point distant 30 m from it, would equal to

Answer:

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20. A combination of capacitors is set-up as shown in the figure. The magnitude of the electric field, due to a point charge $Q$ (having a charge equal to the sum of the charges on the $4 μ F$ and $9 μ F$ capacitors), at a point distant $30 m$ from it, would equal to