Two identical capacitors store different amounts of energy: capacitor A stores 2.4 10-3 J, and capacitor B stores 4.8 10-4 J. The voltage across the plates of capacitor B is 11 V. Find the voltage across the plates of capacitor A.
C=q/v
then q1/v1=q2/V2
solve for V2
To find the voltage across the plates of capacitor A, we can use the formula for the energy stored in a capacitor:
E = (1/2)C(V^2)
Where E is the energy stored, C is the capacitance, and V is the voltage.
Since capacitor B stores 4.8 x 10^-4 J of energy and has a voltage of 11 V, we can write the equation:
4.8 x 10^-4 J = (1/2)C_B(11 V)^2
Simplifying this equation, we have:
4.8 x 10^-4 J = (1/2)C_B(121 V^2)
Now, let's find the capacitance of capacitor B. We can rearrange the equation to solve for C_B:
C_B = (4.8 x 10^-4 J) / [(1/2)(121 V^2)]
C_B = (4.8 x 10^-4 J) / (0.5 x 121 V^2)
C_B = 8 x 10^-6 F
Now that we have the capacitance of capacitor B, we can use the equation to find the voltage across the plates of capacitor A:
2.4 x 10^-3 J = (1/2)C_A(V_A^2)
We can substitute the value of the capacitance calculated from capacitor B:
2.4 x 10^-3 J = (1/2)(8 x 10^-6 F)(V_A^2)
Simplifying this equation:
2.4 x 10^-3 J = 4 x 10^-6 F (V_A^2)
Now solve for V_A:
V_A^2 = (2.4 x 10^-3 J) / (4 x 10^-6 F)
V_A^2 = 600 V^2
Taking the square root of both sides:
V_A = sqrt(600) V
Therefore, the voltage across the plates of capacitor A is approximately 24.5 V.