Charged particle between 2 charged plates

Electricity and Magnetism Level 1

As shown above, a charged particle $A$ with mass $m$ is placed between two parallel charged plates parallel to the ground with the negatively charged plate closer to the ground. The distance between the two plates is $d$ and the electric potential difference between the two plates is $V$ . If the particle is stationary, then what is the magnitude of the quantity of the electric charge of $A?$

Note: $g$ denotes the gravitational acceleration.

\frac{mgd}{V}

\frac{m}{dgV}

\frac{dgV}{m}

\frac{mg}{dV}

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Prakhar Gupta
Jul 8, 2014

In the diagram the charged particle is experiencing two forces, namely:

1>> Electric Force.

2>>Gravitational Force. $F_{g}=mg$ $F_{e}= EQ$ $F_{e}=\dfrac{VQ}{d}$ In Equilibrium the two forces must be equal and opposite. Hence: $mg=\dfrac{VQ}{d}$ $\boxed{Q=\dfrac{mgd}{V}}$

Vishwak Srinivasan
Oct 2, 2014

Assuming we have a uniform electric field,

V = E d,

This implies,

V/d = E,

Electric Force = Force due to Gravity

q E = m g

q V = m g d

q = m g d / V.

Karan Joisher
Jul 6, 2014

As the charge remains stationary the weight of the charge is balanced by electrostatic force F = qE E is potential gradient = V/d F=qV/d mg=qV/d q=mgd/V

Nibir Das
Jul 5, 2014

Electric force = Gravitational Force

qv/d = mg

Sudipan Mallick
Jul 3, 2014

electric field is balanced by downward force applied by the charge particle. so, electric field=V/d therefore QV/d=mg so, we can say q=mgd/V.

Charged particle between 2 charged plates

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