Question 380: Electrostatics & Capacitance

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 the sum of the charges on the $$4$$ $$\mu \,F$$ and $$9$$ $$\mu \,F$$ capacitors), at a point distance $$30$$ $$m$$  from it, would equal :

Accepted Answer

Charge on $${C_1}$$ is $${q_1} = \left[ {\left( {{{12} \over {4 + 12}}} ight) 	imes 8} ight] 	imes 4 = 24\mu c$$
The voltage across $${C_P}$$ is $${V_P} = {4 \over {4 + 12}} 	imes 8 = 2v$$
$$	herefore$$ Voltage across $$9\mu F$$ is also $$2V$$
$$	herefore$$ Charge on $$9\mu F$$ capacitor $$ = 9 	imes 2 = 18\mu C$$
$$	herefore$$ Total charge on $$4\mu F$$ and $$9\mu F$$ $$ = 42\mu c$$
$$	herefore$$ $$E = {{KQ} \over {{r^2}}} = 9 	imes {10^9} 	imes {{42 	imes {{10}^{ - 6}}} \over {30 	imes 30}} = 420N{c^{ - 1}}$$

Answer

$$420N/C$$

Answer

$$480N/C$$

Answer

$$240N/C$$

Answer

$$360N/C$$

Electrostatics & Capacitance

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

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