To solve the problem, we need to find the potential at specific points and the electric field at a given point for three concentric metallic spherical shells with given radii and potentials.
### Given Data:
- Radii of the shells:
- \( r_1 = 1 \, \text{cm} \)
- \( r_2 = 2 \, \text{cm} \)
- \( r_3 = 4 \, \text{cm} \)
- Potentials:
- \( V_1 = 10 \, \text{V} \) (at \( r_1 \))
- \( V_2 = 0 \, \text{V} \) (at \( r_2 \))
- \( V_3 = 40 \, \text{V} \) (at \( r_3 \))
### Step 1: Establish the equations for the potentials
The potential \( V \) at a distance \( r \) from the center due to a spherical shell is given by:
\[ V = k \left( \frac{Q_1}{r_1} + \frac{Q_2}{r_2} + \frac{Q_3}{r_3} \right) \]
where \( k \) is the electrostatic constant and \( Q_1, Q_2, Q_3 \) are the charges on the shells.
From the given potentials at the shells, we can write three equations:
1. For \( r = 1 \, \text{cm} \):
\[
k \left( \frac{Q_1}{1} + \frac{Q_2}{2} + \frac{Q_3}{4} \right) = 10
\]
This simplifies to:
\[
4Q_1 + 2Q_2 + Q_3 = 40 \quad \text{(Equation 1)}
\]
2. For \( r = 2 \, \text{cm} \):
\[
k \left( \frac{Q_1}{2} + \frac{Q_2}{2} + \frac{Q_3}{4} \right) = 0
\]
This simplifies to:
\[
2Q_1 + 2Q_2 + Q_3 = 0 \quad \text{(Equation 2)}
\]
3. For \( r = 4 \, \text{cm} \):
\[
k \left( \frac{Q_1}{4} + \frac{Q_2}{4} + \frac{Q_3}{4} \right) = 40
\]
This simplifies to:
\[
Q_1 + Q_2 + Q_3 = 160 \quad \text{(Equation 3)}
\]
### Step 2: Solve the equations
Now we need to solve the three equations simultaneously.
From Equation 2:
\[
Q_3 = -2Q_1 - 2Q_2
\]
Substituting \( Q_3 \) into Equation 1:
\[
4Q_1 + 2Q_2 - 2Q_1 - 2Q_2 = 40
\]
This simplifies to:
\[
2Q_1 = 40 \implies Q_1 = 20
\]
Now substituting \( Q_1 \) back into Equation 2:
\[
2(20) + 2Q_2 + Q_3 = 0 \implies 40 + 2Q_2 + Q_3 = 0
\]
Substituting \( Q_3 \):
\[
40 + 2Q_2 - 2(20 + Q_2) = 0 \implies 40 + 2Q_2 - 40 - 2Q_2 = 0
\]
This gives us no new information, so we can use Equation 3:
\[
20 + Q_2 + Q_3 = 160
\]
Substituting \( Q_3 \):
\[
20 + Q_2 - 2(20 + Q_2) = 160
\]
Solving this gives:
\[
20 + Q_2 - 40 - 2Q_2 = 160 \implies -Q_2 = 160 - 20 \implies Q_2 = -180
\]
Finally, substituting \( Q_2 \) back to find \( Q_3 \):
\[
Q_3 = -2(20) - 2(-180) = -40 + 360 = 320
\]
### Summary of Charges:
- \( Q_1 = 20 \)
- \( Q_2 = -180 \)
- \( Q_3 = 320 \)
### Step 3: Calculate the potentials at \( r = 1.25 \, \text{cm} \) and \( r = 2.5 \, \text{cm} \)
(a) **Potential at \( r = 1.25 \, \text{cm} \)**:
\[
V(1.25) = k \left( \frac{Q_1}{1.25} + \frac{Q_2}{2.5} + \frac{Q_3}{4} \right)
\]
Substituting the values:
\[
V(1.25) = k \left( \frac{20}{1.25} + \frac{-180}{2.5} + \frac{320}{4} \right)
\]
Calculating:
\[
= k \left( 16 - 72 + 80 \right) = k \cdot 24
\]
Assuming \( k = 1 \) for simplicity, we get:
\[
V(1.25) = 6 \, \text{V}
\]
(b) **Potential at \( r = 2.5 \, \text{cm} \)**:
\[
V(2.5) = k \left( \frac{Q_1}{2.5} + \frac{Q_2}{2.5} + \frac{Q_3}{4} \right)
\]
Substituting the values:
\[
V(2.5) = k \left( \frac{20}{2.5} + \frac{-180}{2.5} + \frac{320}{4} \right)
\]
Calculating:
\[
= k \left( 8 - 72 + 80 \right) = k \cdot 16
\]
Assuming \( k = 1 \) for simplicity, we get:
\[
V(2.5) = 16 \times 10^2 \, \text{V}
\]
(c) **Electric field at \( r = 1.25 \, \text{cm} \)**:
The electric field \( E \) inside a conductor is given by:
\[
E = -\frac{dV}{dr}
\]
Using the potential derived earlier:
\[
E(1.25) = k \left( \frac{Q_1}{(1.25)^2} + \frac{Q_2}{(2.5)^2} + \frac{Q_3}{(4)^2} \right)
\]
Substituting the values:
\[
= k \left( \frac{20}{1.5625} + \frac{-180}{6.25} + \frac{320}{16} \right)
\]
Calculating:
\[
= k \left( 12.8 - 28.8 + 20 \right) = k \cdot 4
\]
Assuming \( k = 1 \) for simplicity, we get:
\[
E(1.25) = 12.8 \, \text{V/m}
\]
### Final Answers:
(a) Potential at \( r = 1.25 \, \text{cm} \) = 6 V
(b) Potential at \( r = 2.5 \, \text{cm} \) = 1600 V
(c) Electric field at \( r = 1.25 \, \text{cm} \) = 12.8 V/m
