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Question: Electrostatic energy of \(3.5 \times {10^{ - 4}}\) J is stored in a capacitor at 700V. What is the c...

Electrostatic energy of 3.5×10−43.5 \times {10^{ - 4}} J is stored in a capacitor at 700V. What is the charge on the capacitor?

Explanation

Solution

As question says Electrostatic energy of 3.5×10−43.5 \times {10^{ - 4}} J and given voltage 700V. so, we have to apply voltage and energy related formulas, by putting these values we easily calculate the charge on the capacitor.

Formula used:
Q=2uVQ = \dfrac{{2u}}{V}

Complete Step by step solution:
We know that formula
u=12QVu = \dfrac{1}{2}QV
u is the electrostatic energy.
Q is the charge.
V is the voltage.

Given values are u=3.5×10−4u = 3.5 \times {10^{ - 4}}, V=700V, Q have to find.

From the formula we get.
⇒Q=2uV\Rightarrow Q = \dfrac{{2u}}{V}

Subtitue given values in the above equation
⇒Q=2×3.5×10−4700\Rightarrow Q = \dfrac{{2 \times 3.5 \times {{10}^{ - 4}}}}{{700}}

Solve above equation we get
∴Q=10−6C\therefore Q = {10^{ - 6}}C

Hence, charge on the capacitor is 10−6C{10^{ - 6}}C.

Additional information:
∙\bullet Capacitor is a device which stores electrical energy in the form of an electrical field.
∙\bullet Commonly asked question in the viva is what is capacitance, we have to answer the effect of capacitor is known as capacitance.
∙\bullet Electrostatic means electric charges at rest. Electrodynamics means motion of electric charges.
∙\bullet Electrostatic may involve buildup of charge on the surface of material due to contact with other surfaces.
∙\bullet Voltage is nothing but electric potential difference. The dimension of the voltage is ML2T - 3I - 1{\text{M}}{{\text{L}}^{\text{2}}}{{\text{T}}^{{\text{ - 3}}}}{{\text{I}}^{{\text{ - 1}}}} and its unit is volt which is denoted as V.
∙\bullet Electric current is proportional to voltage and inversely proportional to resistance which is stated in a ohm's law.
∙\bullet Resistance is measured in ohms, resistance means the measurement of opposition to current flow.

Note:
We know that here one faraday capacitor at a voltage of 1 volt stores one-coulomb is equivalent to 6.25×10186.25 \times {10^{18}} electrons and a current of 1A shows rate of 1 coulomb each second, hence there is a capacitor of one faraday at 1 volt can store one ampere second electron.