A parallel-plate capacitor of surface area of 10cm2 and plate separation distance (d) of 3mm is connected to a 15V battery as shown in the following figure. Find the work done to decrease the plate separation distance to 1mm while the capacitor is connected to the battery.
Question
A parallel-plate capacitor of surface area of 10cm2 and plate separation distance (d) of 3mm is connected to a 15V battery as shown in the following figure. Find the work done to decrease the plate separation distance to 1mm while the capacitor is connected to the battery.
Solution 1
To solve this problem, we need to understand that when a capacitor is connected to a battery, the voltage across the capacitor remains constant, even if the plate separation distance changes. This is because the battery continuously supplies or removes charge to keep the voltage constant.
Step 1: Calculate the initial capacitance (C1) using the formula C = ε0 * (A/d), where ε0 is the permittivity of free space (8.85 x 10^-12 F/m), A is the area of one of the plates (convert from cm^2 to m^2), and d is the separation distance (convert from mm to m).
Step 2: Calculate the final capacitance (C2) using the same formula, but with the new separation distance.
Step 3: The work done by the battery to change the capacitance while keeping the voltage constant is given by the formula W = 0.5 * V^2 * (C2 - C1), where V is the voltage of the battery.
By substituting the given values into these formulas, you can calculate the work done.
Solution 2
To solve this problem, we need to understand that when a capacitor is connected to a battery, the voltage across the capacitor remains constant, but the capacitance and charge stored on the capacitor can change.
-
First, we need to calculate the initial capacitance of the capacitor. The formula for the capacitance (C) of a parallel-plate capacitor is C = ε0 * (A/d), where ε0 is the permittivity of free space (8.85 x 10^-12 F/m), A is the area of one of the plates, and d is the distance between the plates.
Given that A = 10 cm^2 = 10^-3 m^2 (since 1 cm^2 = 10^-4 m^2) and d = 3 mm = 3 x 10^-3 m, we can substitute these values into the formula to get:
C1 = ε0 * (A/d) = 8.85 x 10^-12 F/m * (10^-3 m^2 / 3 x 10^-3 m) = 2.95 x 10^-12 F.
-
Next, we calculate the charge (Q) stored on the capacitor using the formula Q = CV, where V is the voltage across the capacitor. Given that V = 15 V, we can substitute the values to get:
Q = C1 * V = 2.95 x 10^-12 F * 15 V = 44.25 x 10^-12 C.
-
Then, we calculate the new capacitance (C2) when the plate separation distance is decreased to 1 mm = 1 x 10^-3 m. Using the formula for capacitance, we get:
C2 = ε0 * (A/d) = 8.85 x 10^-12 F/m * (10^-3 m^2 / 1 x 10^-3 m) = 8.85 x 10^-12 F.
-
The work done (W) to decrease the plate separation distance while the capacitor is connected to the battery can be calculated using the formula W = 0.5 * C * V^2. However, since the voltage remains constant, the work done is actually the change in the energy stored in the capacitor, which is given by ΔW = 0.5 * ΔC * V^2.
Substituting the values, we get:
ΔW = 0.5 * (C2 - C1) * V^2 = 0.5 * (8.85 x 10^-12 F - 2.95 x 10^-12 F) * (15 V)^2 = 0.5 * 5.9 x 10^-12 F * 225 V^2 = 0.66 x 10^-12 J = 0.66 pJ.
So, the work done to decrease the plate separation distance to 1mm while the capacitor is connected to the battery is approximately 0.66 picojoules.
Similar Questions
A parallel plate capacitor filled with mica having εr = 5 is connected to a 10 V battery. The area of the parallel plate is 6 m^2 and separation distance is 6 mm. Find the capacitance.*1 point4.43 × 10^-8 F4.43 × 10^-9 F44.3 × 10^-8 F443 × 10^-9 F
A capacitor is constructed with two parallel metal plates each with an area of 0.74 m2 and separated by d = 0.80 cm. The two plates are connected to a 4.0-volt battery. The current continues until a charge of magnitude Q accumulates on each of the oppositely charged plates.Find the electric field in the region between the two plates. V/mFind the charge Q. CFind the capacitance of the parallel plates.
Solve the given problem below. Submit your illustration and solutions as PDF format as an attachment to your final answer (60 points). A parallel plate capacitor is constructed of two (2) square conducting plates with side length l = 10.0 cm. The distance between the plates is d = 0.250 cm. A dielectric with a dielectric constant k = 15.0 and thickness 0.250 cm is inserted between the plates. The dielectric is 10.0 cm wide and is 5.0 cm long.Provide an illustration that best represents the scenario (20 points).Derive and provide the formula that shall best address the given scenario (20 points).Determine the capacitance of this system (20 points).
A capacitor of plate area 0.01 m2 and plate separation 0.05 mm is charged by a battery of potential 10 V then the charge on positive plate will beChoose answer: 0.18 μC 0.17 nC 8.8 nC 17.7 nC
8 A parallel plate capacitor of capacity 100𝜇F is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are brought closer so that the distance between them becomes half the original distance, the additional energy given by the battery to the capacitor in joules is: a) 1.25 × 10ିଷ b) 0.125 × 10ିଷ c) 12.5 × 10ିଷ d) 125 × 10ିଷ 9 Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 volt and the average resistance per km is 0.5𝛺 . The power loss in the wires is: a) 19.2 kW b) 12.2J c) 12.2kW d) 19.2 J 10 The resistivity of the potentiometer wire is 10ି 𝛺 m. Its area of cross - section is 10 ି m ଶ . When a current I = 0.1. A flow through the wire, its potential gradient is: a) 10ିସ Vm ିଵ b) 10Vmିଵ c) 10ିଶ Vm ିଵ d) 0.1 Vm ିଵ 11 A set of n equal resistors, of value R each, are connected in series to a battery of emf 𝜀 and internal resistance R. The current drawn is I. Now, the n resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of n is: a) 10 b) 11 c) 20 d) 9 12 According to Kirchhoff’s Junction Rule, a) none of these b) At any junction of circuit elements, the sum of currents entering and leaving the junction must be positive.c) At any junction of circuit elements, the sum of currents entering the junction must equal the sum of currents leaving it. d) At any junction of circuit elements, the sum of currents entering and leaving the junction must be negative. 13 A galvanometer having a resistance of 8𝛺 is shunted by a wire of resistance 2 𝛺 . If the total current is 1 A, the part of it passing through the shunt will be: a) 0.8 A b) 0.3 A c) 0.5 A d) 1.2 A 14 A circular coil A has a radius a and the current flowing in it is I. Another circular coil B has a radius 2a and if 2lis the current flowing through it, then the magnetic fields at the centre of the circular coils are in the ratio of: a) 3 : 1 b) 4 : 1 c) 1 : 1 d) 2 : 1 15 A galvanometer of resistance 20𝛺 shows a deflection of 10 divisions when a current of 1 mA is passed through it. If a shunt of 4 𝛺 is connected and there are 50 divisions on the scale, the range of the galvanometer is: a) 40 mA b) 30 mA c) 25 mA d) 35 mA 16 According to Gauss’s law for magnetism, a) ∮ 𝐵ሬ⃗. 𝑑𝑠 ሬሬሬሬ⃗ = 0 b) ∫ 𝐵ሬ⃗. 𝑑𝑠 ሬሬሬሬ⃗ = 0 c) ∮ 𝐵ሬ⃗. 𝑑𝑠 ሬሬሬሬ⃗ = 𝜇 d) ∫ 𝐵ሬ⃗. 𝑑𝑠 ሬሬሬሬ⃗ = 𝜇 17 A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2A? a) 3.48 T b) 5.48 T c) 4.08 T d) 4.48 T 18 Magnetic field due to a magnet 2 cm long having a pole strength of 100 Am at a point 10 cm from each pole is a) 8 × 10ିସ𝑇 b) 4 × 10ିହ𝑇 c) 2 × 10ିହ𝑇 d) 2 × 10ିସ𝑇
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.