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3 years ago

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A capacitor consists of two parallel plates, each with an area of 16.0 cm2 , separated by a distance of 0.150 cm . The material

that fills the volume between the plates has a dielectric constant of 6.00. The plates of the capacitor are connected to a 400 V battery.

1 answer:

Alchen [17]3 years ago

7 0

Answer:

Explanation:

Given

area of capacitor $A=16\ cm^2$

Distance between Capacitor $d=0.15\ cm$

dielectric constant $k=6$

Voltage applied $V=400\ V$

Capacitance C is given by $C=k\frac{\epsilon A}{d}$

$C=6\times \frac{8.85\times 10^{-12}\times 16\times 10^{-4}}{0.15\times 10^{-2}}$

$C=56.64\times 10^{-12}\ F$

Charge on Plates

$Q=C\times V$

$Q=56.64\times 10^{-12}\times 400$

$Q=22,656\times 10^{-12}$

$Q=22.656\ nC$

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Two spherical shells have a common center. A -1.50 × 10-6 C charge is spread uniformly over the inner shell, which has a radius

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Answer:

a) At 0.20 m, the magnitude of the field is 675.0 kV

The direction of the field is acting outwards from the center of the charged spheres

b) At 0.10 m, the magnitude of the field is 135 kV

The direction is acting outwards from the center of the charged spheres

c) At 0.025 m

The magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres

Explanation:

The charged spherical shell parameters are;

The charge on the inner sphere, q₁ = -1.50 × 10⁻⁶ C

The radius of the inner shell, R₁ = 0.050 m

The charge on the outer sphere, q₂ = +4.50 × 10⁻⁶ C

The radius of the outer shell, R₂ = 0.15 m

Let 'r', represent the distance at which the electric field is measured, the following relationships can be obtained;

When r < R₁ < R₂,

$V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )$

When R₁ < r < R₂,

$V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )$

When R₁ < R₂ < r,

$V = k \cdot \left( \dfrac{q_1 + q_2}{r^2} \right )$

a) When r = 0.20 m, we have;

R₁ < R₂ < r, therefore

$V = k \cdot \left( \dfrac{q_1 + q_2}{r^2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} + 4.50\times 10^{-6} }{0.20^2} \right ) = 675.0 \ kV$

Therefore, the magnitude of the field, V = 675.0 kV

The direction of the field is outwards

b) When r = 0.10 m, we have;

When R₁ < r < R₂, therefore;

$V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} }{0.10} + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = 135 \ kV$

Therefore, the magnitude of the field, V = 135 kV

The direction of the field is outwards from the center

c) When r = 0.025 m, we have;

When r < R₁ < R₂, therefore;

$V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )$

By plugging in the values, we get;

$V = 9 \times 10^9 \times \left( \dfrac{-1.50 \times 10^{-6} }{0.05} + \dfrac{4.50\times 10^{-6}}{0.15} \right ) = -270 \ kV$

Therefore, the magnitude of the field, V = -270.0 kV

The direction of the field is inwards, towards the center of the charged spheres.

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Answer:

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Explanation:

It is given that,

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Initial velocity at point A, $u=4.52\times 10^5\ m/s$

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

$v^2=u^2+2as$ ........(1)

a is the acceleration, $a=\dfrac{F}{m}$

We know that electric force, F = qE

$a=\dfrac{qE}{m}$

Use above equation in equation (1) as:

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$v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m$

v = 647302.09 m/s

or

$v=6.45\times 10^5\ m/s$

So, the final velocity of the electron when it reaches point B is $6.45\times 10^5\ m/s$ . Hence, this is the required solution.

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