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 distant $30\, m$ from it, would equal :

• A. 240 N/C
• B. 360 N/C
• C. 420 N/C
• D. 480 N/C

Answer 1

The Correct Option is C

 To solve this problem, we need to find the magnitude of the electric field due to a point charge at a certain distance. The question provides a point charge \( Q \), which is the sum of charges on the \( 4 \, \mu F \) and \( 9 \, \mu F \) capacitors.

Let us denote the charges on these capacitors as \( Q_1 \) and \( Q_2 \) respectively. Assuming the capacitors are connected in such a way that they have the same potential difference, we can use the formula \( Q = CV \), where \( C \) is capacitance and \( V \) is voltage.

The charge on the capacitors can be written as:

• \( Q_1 = 4 \, \mu F \times V \)
• \( Q_2 = 9 \, \mu F \times V \)

Therefore, the total charge \( Q \) is given by:

\(Q = Q_1 + Q_2 = (4 + 9) \, \mu F \times V = 13 \, \mu F \times V\)

The electric field \( E \) due to a point charge \( Q \) at a distance \( r \) is given by the formula:

\(E = \frac{k \times Q}{r^2}\)

where \( k = 8.99 \times 10^9 \, Nm^2/C^2 \) is the Coulomb's constant.

Now, substituting the values given in the question:

• \(Q = 13 \, \mu C = 13 \times 10^{-6} \, C\)
• \(r = 30 \, m\)

Plug these into the formula for the electric field:

\(E = \frac{8.99 \times 10^9 \times 13 \times 10^{-6}}{30^2}\)

Calculate the value:

\(E = \frac{8.99 \times 10^9 \times 13 \times 10^{-6}}{900}\)

Simplifying the calculation gives:

\(E = \frac{116870}{900} = 420 \, N/C\)

Therefore, the magnitude of the electric field due to the point charge at a distance of 30 m is 420 N/C.

Thus, the correct answer is:

420 N/C