Physics, 08.02.2021 22:20, erykp17

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 of 0.050 m. A +4.50 × 10-6 C charge is spread uniformly over the outer shell, which has a radius of 0.15 m. Find the magnitude and direction of the electric field at a distance (measured from the common center) of. a: 0.20m

b: 0.10m

c: 0.025m

Answers: 2

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Answers

answered: stefkellum58

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.

answered: Guest

sleep paralysis is a parasomnia, events associated with sleep.

sleep paralysis happens when you are in the middle of waking up or falling asleep. it is the feeling of being conscious but unable to move. it definitely involves hallucinations.

i had a friend that had sleep paralysis and he told me a story one time where he was laying in bed and he couldn't move his body no matter how hard he tried and there was someone standing in the corner of his room coming towards him. he couldn't move and you can't "wake up" because you are already conscious. sometimes sleep paralysis can be triggered by frequent waking up during the night based off of multiple trips to the bathroom, or if you had ever been abused, sleep paralysis can start happening from things like that.

answered: Guest

ingenuity says that the answer is 2.7 m.

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Two spherical shells have a common center. A -1.50 × 10-6 C charge is spread uniformly over the inne...