What force will a charge of 1C feel when it is in the electric field with a strength of 100 V/m?

What force will a charge of 1C feel when it is in the electric field with a strength of 100 V/m?

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How much kinetic energy will a particle with a charge of 10 mC gain while accelerating throgh a potential difference of 1kV?

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Two planes are placed parallel in a vacuum; there is a potential difference of 10kV between them. The distance between the planes is 10cm. Calculate the electrical field strength between these two planes

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Calculate the force between two charged particles with charges of Q1 = 1mC and Q2 = 2mC placed at the distnace of 2m in vacuum.

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\(F_{coulomb}={kQ_qQ_2\over r^2}\approx {8.99\times10^9\times 0.001\times0.002\over4}=4495N\)

What is the capacitance of a capacitor which can store 24mC of charge when the potential difference across it is 12V?

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\(C = Q/V =24\times 10^{-3}/12 = 2mF\)

What is the energy stored on a 400 \(\mu F\) capacitor which has 2mC of charge on it?

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\(E = {Q^2\over 2C} ={(2\times10^{-3})^2\over2 \times 400\times10^{-6}} = 0.005J\)

Select the correct expression for time constant of RC-circuit.

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\(\tau = RC\). Time constant shows how long it takes for a resistor-capacitor system to discharge.

А 0.04 F capacitor is fully charged by a 5V supply and is then connected to discharge through a \(800\Omega\) resistor. How much charge remains on the capacitor after 20 seconds?

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Intial charge of the capacitor is \(Q_0=CV=0.2C\).

The function describing remaining charge is \(Q(t)=Q_0e^{-t/RC}=0.2\times e^{-20/{800\times0.04}}\approx0.11C\).

А 0.02 F capacitor is fully charged by a 5 V supply and is then connected to discharge through a \(800\Omega\) resistor. What is the potential difference on the capacitor after 10 seconds?

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\(V =V_0\times e^{-t/RC} = 5\times e^{-10/800\times 0.02}\approx2.68V\)

А 0.02 F capacitor is fully charged by a 5 V supply and is then connected to discharge through a \(800\Omega \) resistor. What is the discharge current after 10 seconds?

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Initial current is \(I_0 =V_0/R= 5/800=6.25mA\).

\(I = I_0\times e^{-t/RC}=6.25\times e^{-10/0.02\times 800}\approx3.35mA\)