The energy held in a capacitor can be calculated using
, where C represents capacitance and V denotes the voltage applied across the capacitor.
We will compute the energy for the first capacitor:
Next, let's find the energy for the second capacitor:
Thus, the combined energy stored in both capacitors amounts to
Answer:
The minimum resistance value is 
Explanation:
According to the question, we have:
The voltage given as 
The internal resistance as 
The goal here is to find the minimum resistance for the voltmeter such that its reading is within 1.0% of the battery's emf.
This means we require voltmeter resistance such that:
V = (100% - 1%) of E
Where E is the battery's e.m.f. and V is the voltmeter reading.
So, V = 99% of E = 0.99 E = 7.425.
In general, we have:
E = V + ir
where ir denotes the internal voltage drop across the voltmeter, and V is the reading from the voltmeter.
By rearranging, we get:
Since the current remains constant throughout the circuit:
where 
is the voltmeter resistance value.
Hence, 
Response:
Magnitude of the electrostatic force acting on the +32 µC charge, 
Clarification:
Let q₁ = +32 µC, located at x₁ = 0
q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m
q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m
Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).
Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the force on q₁ due to q₃).
The resultant electrostatic force on the 32 µC charge is 
