Question

19.
Two parallel plates 6.0 x 102m long are separated by
25x10 m and have a potential difference of 850 V.
Point P is located midway between the two plates as
shown below
OV
2.5x10 m
850 V
- 6.0x10 m
What is the magnitude of the electric field at point P?​

Answer 1

Answer:

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Answer 2

Answer: The correct question is " Two parallel plates $6.0 X 10^{-2}$ m long are separated by $2.5X10^{-2}$ m and have a potential difference of 850 V. Point P is located midway between the two plates. What is the magnitude of the electric field at point P?​

The answer is: 3.4 kV/m

Explanation:

Step 1: The given data:

There are two parallel plates of dimension $6.0 X 10^{-2}$

Distance between plates is, d = $2.5X10^{-2} m$

Potential difference between the plate is, V = 850 V

Step 2: Find the Capacitance of the plate.

We know capacitance C for parallel plate capacitor with air in between is given by , C = $\frac{A\epsilon_0}{d}$

Where A is area of the plate = dimension X dimension = $6X10^{-2}X6X10^{-2}=3.6X10^{-3}$ $m^2$

$\epsilon_0 = 8.85X10^{-12}$ is permittivity of free space

d is distance between the plates.

Therefore,

C = $\frac{3.6X10^{-3}X8.85X10^{-12}}{2.5X10^{-2}}$

C = $12.74X10^{-13}$F = 1.274 pF

Step 3: Find the charge on the plate

We know, Q = CV

where Q is charge on the plate.

Q = 1.274pF X 850V = $1.1 X 10^{-9} As$

Step 4: Find the surface charge density $\sigma$ of the plate

We know that,

$\sigma = \frac{Q}{A} \\\sigma = \frac{1.08X10^{-9}}{3.6X10^{-3}}\\\sigma = 3X10^{-7}Asm^{-2}$

Step 5: Find the Electric field E between the plates.

For two parallel plates of opposite charges field between the plate(everywhere) remains the same.

Therefore, E = $\frac{\sigma}{\epsilon_0}$

E = $\frac{3X10^{-7}}{8.85X10^_-12}}$

E = $3.4 X 10^3 Vm^{-1}$

Therefore, The electric field between plates is 3.4 kV/m

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