Question

Two parallel square metal plates that are 1.6 cm apart and 24 cm on each side carry equal but opposite charges uniformly spread out over their facing surfaces. How many excess electrons are on the negative surface if the electric field between the plates has a magnitude of 18,000 N/C

Answer 1

The amount of extra electrons present on the negative surface is

57.2 x $10^{10}$.

Distance =1.6 cm

Side = 24 cm

Electric field = 18000 N/C

Calculating the capacitance in the metal plates is necessary.

Using the capacitance formula

$C=\frac{e_{0}A }{d}$

Putting the value

C = 8.85 x $10^{-12}$x (24 x $10^{-2}$$)^{2}$/1.6 x $10^{-2}$

C = 0.318 x $10^{-10}$ F

Calculation of potential

V = Ed

V = 18000 x 1.6 x $10^{-2}$ V

V = 288 V

Calculation of charge

Q = CV

Q = 0.318 x $10^{-10}$ x 288

Q = 91.54 x $10^{-10}$ C

Charge on the both the plates

Q = +91.54 x $10^{-10}$

Q = - 91.54 x $10^{-10}$

Calculation of excess electrons on the negative surface:

n = q/e

n = 91.54 x $10^{-10}$/ 1.6 x $10^{-19}$

n = 57.2 x $10^{10}$ electrons

Hence, the number of excess electrons on the negative surface is

57.2 x $10^{10}$.

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brainly.com/question/14746225

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