Question

 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 sum of capacitance  on $4\mu C$ and $9\mu C$ capacitors), at a point 30 m form it, would equal

A. 250 N/C

B. 360 N/C

C. 420 N/C

D. 480 N/C

Answer 1

Hint: While solving this question, try to solve each part one by one. By doing so will help make the diagram more simplified. Remember the additional rule of capacitors. Capacitors in parallel are added directly while those in series have to be added inversely. 

Complete answer:

Let us solve the question by breaking the diagram 

For the capacitors with capacitance  $4\mu C$, $3\mu C$ and $9\mu C$, the system will be:

Since the capacitors with capacitance  $3\mu C$ and $9\mu C$are in parallel system,

The net charge will be the sum of their individual capacitance, i.e.,

$\Rightarrow 3\mu C+~~9\mu C=12\mu C$

So now, the resultant system will be

Now,

The capacitors shown in the above figure will be in series

So, the net charge will be the sum of inverse of the capacitance  

Hence, we have

$\Rightarrow {{(\dfrac{1}{4}+\dfrac{1}{12})}^{-1}}={{(\dfrac{1}{3})}^{-1}}=3\mu C$ 

Now, 

Total charge will be,

\[\Rightarrow Q=CV=3\times 8=24\mu C\]

This is the charge distributed by the direct ratio of capacitors $3\mu C$ and $9\mu C$

So,

 \[\Rightarrow {{Q}_{9\mu F}}=\dfrac{24\times 9}{3+3}=18\mu C\] 

Now, 

\[{{Q}_{9\mu F}}+{{Q}_{4\mu F}}=18+24=42\mu C\] 

So, 

The magnitude of the electric field due to a point charge \[42\mu C\] at a point 30 m form it, would equal 

\[\Rightarrow E=\dfrac{kQ}{{{r}^{2}}}\] 

\[\Rightarrow E=\dfrac{9\times {{10}^{9}}\times 42\times {{10}^{-6}}}{900}\] 

\[\Rightarrow E=420N/C\]

So, The magnitude of the electric field due to a point charge Q (having a charge equal to the sum of capacitance  on $4\mu C$ and $9\mu C$ capacitors), at a point 30 m from it, would equal 420 N/C.

So, the correct answer is “Option C”.

Note:

Remember the addition rule for the capacitors are opposite of those for the resistances. In the case of resistance, the resistance in parallel are added inversely while those in series are added directly. Also, always remember to change the units into the ones that are given in the option for finding the correct answer. For example in this question, the value that was given for the capacitors were on the micro scale. So, 

$1\mu C={{10}^{-6}}C$