Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
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,
[tex]V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )[/tex]
When R₁ < r < R₂,
[tex]V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )[/tex]
When R₁ < R₂ < r,
[tex]V = k \cdot \left( \dfrac{q_1 + q_2}{r^2} \right )[/tex]
a) When r = 0.20 m, we have;
R₁ < R₂ < r, therefore
[tex]V = k \cdot \left( \dfrac{q_1 + q_2}{r^2} \right )[/tex]
By plugging in the values, we get;
[tex]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[/tex]
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;
[tex]V = k \cdot \left( \dfrac{q_1}{r} + \dfrac{q_2}{R_2} \right )[/tex]
By plugging in the values, we get;
[tex]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[/tex]
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;
[tex]V = k \cdot \left( \dfrac{q_1}{R_1} + \dfrac{q_2}{R_2} \right )[/tex]
By plugging in the values, we get;
[tex]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[/tex]
Therefore, the magnitude of the field, V = -270.0 kV
The direction of the field is inwards, towards the center of the charged spheres.
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
