JEE Main 20163 April
- Exam
- JEE Main
- Year
- 2016
- Date
- 3 April
- Subject
- Physics
- Chapter
- Capacitance
JEE Main 2016 · 3 April · Physics · Capacitance
A combination of capacitors is set up as shown in the figure. The...
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:
Options
- A420 N/C
- B480 N/C
- C240 N/C
- D360 N/C
Correct answer
A. 420 N/C
Step-by-step solution
Ceq = 5μF<br /><br />Potentialof4μF = 6volt
∴ charge\text{on}4μF\left({\text{q}}_{\text{4}}\right) = 24μC
Potential of9μF = 2volt
∴ charge\text{on}9μF\left({\text{q}}_{9}\right) = 18μC
Totalcharge ( q ) = 42μCE = \dfrac{kq}{{r}^{2}} = \dfrac{9 × {10}^{9} × 42 × {10}^{- 6}}{900} = 420N / C