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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 µF and 9 µF capacitors), at a point distant 30 m from it, would equal :

Option 1)
240 N/C

Option 2)
360 N/C

Option 3)
420 N/C

Option 4)
480 N/C

Answers (2)

As we learnt in

Series Grouping -

$\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\cdots$

- wherein

Parallel Grouping -

$C_{eq}=C_{1}+C_{2}+\cdots$

- wherein

Electric Field Intensity -

$\vec{E}=\frac{\vec{F}}{q_{0}}=\frac{kQ}{r^{2}}$

- wherein

$3 \mu F\:and\: 9\mu F$ are in parallel combinations so their equivalent capacitance is

$= 3+9 = 12 \mu f$

Now $4 \mu F\:and\: 12\mu F$ are in series sso their equivalent capacitance

$= \frac{4 \times 12}{16} = 3\mu F$

Charge on $3 \mu F$ = $(3 \mu f) \times (8V) = 24 \mu C$

$\therefore$ Charge on $4 \mu F\:and\: 12\mu F$ are same

$(24 \mu C)$ as they are in series

Charge on $9 \mu F= \frac{9}{9+3} \times 24 \mu F = 18 \mu C$

Now required charge on Q = charge on $4 \mu F$ + charge on $9 \mu F$

$\therefore Q = [24+18] \mu C= 42 \mu C$

$\therefore E = \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2} \implies E = 9 \times 10^9 \times \frac{42 \times 10^{-6}}{(30)^2}= 420 N C^{-1}$

Option 1)

240 N/C

Incorrect

Option 2)

360 N/C

Incorrect

Option 3)

420 N/C

Correct

Option 4)

480 N/C

Incorrect

Posted by

Vakul

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