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

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