A solid core carrying excess charge, Q = 46.4 μC, is located at the center of concentric spherical shells. The two spheres are joined by a hollow plastic tube that starts at the radius of the small sphere, r = 17.5 cm, and ends at the radius of the large one, R = 38.5 cm. When a charge, q = 6.9 µC, is placed at the inner end of the tube and released from rest, it accelerates straight through the tube and out the other end as shown. 1 -kQq ( ² - 1/2) R (a) How much did the electric potential energy change as the particle moved from one end of the tube to the other? Write your answer symbolically, in terms of Q, q, r, R and constants (as needed), then calculate its numeric value. AUε = = 8.97 R x J ОГ (b) How much electric potential energy will be left when the little charge is 73.5 cm from the core? UE = 7.48 x J
A solid core carrying excess charge, Q = 46.4 μC, is located at the center of concentric spherical shells. The two spheres are joined by a hollow plastic tube that starts at the radius of the small sphere, r = 17.5 cm, and ends at the radius of the large one, R = 38.5 cm. When a charge, q = 6.9 µC, is placed at the inner end of the tube and released from rest, it accelerates straight through the tube and out the other end as shown. 1 -kQq ( ² - 1/2) R (a) How much did the electric potential energy change as the particle moved from one end of the tube to the other? Write your answer symbolically, in terms of Q, q, r, R and constants (as needed), then calculate its numeric value. AUε = = 8.97 R x J ОГ (b) How much electric potential energy will be left when the little charge is 73.5 cm from the core? UE = 7.48 x J
Related questions
Question
Transcribed Image Text:A solid core carrying excess charge, Q = 46.4 μC, is
located at the center of concentric spherical shells.
The two spheres are joined by a hollow plastic tube that
starts at the radius of the small sphere, r = 17.5 cm, and
ends at the radius of the large one, R = 38.5 cm.
When a charge, q = 6.9 μC, is placed at the inner end of
the tube and released from rest, it accelerates straight
through the tube and out the other end as shown.
- kog ( ² = − 1 / 2 )
kQq
R
(a) How much did the electric potential energy change as the particle moved from one end of the tube to the other?
Write your answer symbolically, in terms of Q, q, r, R and constants (as needed), then calculate its numeric value.
AUE=
= 8.97
X
R
x J
(b) How much electric potential energy will be left when the little charge is 73.5 cm from the core?
UE = 7.48
J
9
Expert Solution
Step 1
We will answer the question using expression for electric potential due to a source charge, and expression for potential energy when a test charge is brought near that source charge. The steps are as follows.
Step by step
Solved in 4 steps
