Question

The two long plates are separated by a distance of $15.0 cm$ . The bottom plate is kept at a potential of $- 250 V$ and the top at $+ 250 V$ . A charge $- 2.00 μC$ is placed at a point $3.00 cm$ from the bottom plate.

The charge is then moved vertically up to a point $3.00 cm$ from the top plate.

What is the electric potential energy of the charge now $?$

Accepted Answer

The potential energy at a distance $d$ from the bottom plate is

$V = V_{b o t t o m} + E \times d$

Here, $V_{b o t t o m}$ is the electric potential of the bottom plate and $E$ is the electric field between two plates.

Since, the electric field between two plates is $\frac{∆ V}{∆ r}$ . Here, $∆ V$ is the potential difference between two plates and $∆ r$ is the separation between the plates.

$E = \frac{250 - (- 250)}{15 \times 10^{- 2}} = \frac{500}{0.15} N C^{- 1}$

The charge is now at a distance of $12.0 cm$ from the bottom plate. So, the electric potential at this point is

$V = - 250 + \frac{500}{0.15} \times 12 \times 10^{- 2}$

$V = 150 V$

Here, $V_{b o t t o m}$ is the electric field of the bottom plate and $E$ is the electric potential between two plates. Since, the electric field between two plates is $\frac{∆ V}{∆ r}$ . Here, $∆ V$ is the potential difference between two plates and $∆ r$ is the separation between the plates.

$E = \frac{250 - (- 250)}{15 \times 10^{- 2}} = \frac{500}{0.15} N C^{- 1}$

So, $V = - 250 + \frac{500}{0.15} \times 3 \times 10^{- 2}$

$V = - 150 V$

So, the electric potential energy is $E_{P} = V \times q$

$E_{P} = 150 \times - 2.00 \times 10^{- 6} = - 3 \times 10^{- 4} J$

K A Tsokos Solutions for Chapter: Fields (HL), Exercise 1: Test yourself

K A Tsokos Physics Solutions for Exercise - K A Tsokos Solutions for Chapter: Fields (HL), Exercise 1: Test yourself

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