What Capacitance is required to store an energy of 10 kWh at a potential difference of 1000 volt.

Hey Em_Legend, good to see you again. Here are the solutions to your physics problems: Question 1: What Capacitance is required to store an energy of 10 kWh at a potential difference of 1000 volt. Step 1: Convert the energy from kilowatt-hours (kWh) to Joules (J). 1 kWh = 3.6 × 10^6 J U = 10 kWh × 3.6 × 10^6 J/kWh = 3.6 × 10^7 J Step 2: Use the formula for energy stored in a capacitor, U = (1)/(2)CV^2, and rearrange it to solve for capacitance C. C = (2U)/(V^2) Step 3: Substitute the given values into the formula and calculate C. C = 2 × 3.6 × 10^7 J(1000 V)^2 C = 7.2 × 10^7 J1 × 10^6 V^2 C = 72 F The required capacitance is 72 F. Question 2: How much energy is store in 1m³ (cubic meter) of air do to the fair weather electric field of magnitude 15 volt per meter. Step 1: Identify the given values and the permittivity of free space (_0). Given: Volume Vol = 1 m^3, Electric field E = 15 V/m. The permittivity of free space (which is approximately the permittivity of air) is _0 = 8.854 × 10^-12 F/m. Step 2: Calculate the energy density (u) using the formula u = (1)/(2)_0 E^2. u = (1)/(2) × (8.854 × 10^-12 F/m) × (15 V/m)^2 u = (1)/(2) × 8.854 × 10^-12 F/m × 225 V^2/m^2 u = 9.96075 × 10^-10 J/m^3 Step 3: Calculate the total energy stored (U) by multiplying the energy density by the volume. U = u × Vol U = (9.96075 × 10^-10 J/m^3) × (1 m^3) U = 9.96075 × 10^-10 J The energy stored is 9.96 × 10^-10 J. Question 3: Calculate the Capacitance of a Capacitors whose Area 1m² and distance between the plate is 8.8 x 10⁻¹² m dielectric between the plate is the factors of 5. Step 1: Identify the given values and the permittivity of free space (_0). Given: Area A = 1 m^2, Distance d = 8.8 × 10^-12 m, Dielectric constant = 5. The permittivity of free space is _0 = 8.854 × 10^-12 F/m. Step 2: Use the formula for the capacitance of a parallel plate capacitor with a dielectric: C = ( _0 A)/(d). Step 3: Substitute the values into the formula and calculate C. C = 5 × (8.854 × 10^-12 F/m) × (1 m^2)8.8 × 10^-12 m C = 44.27 × 10^-12 F8.8 × 10^-12 C = 5.03068 F The capacitance is 5.03 F. Send me the next one 📸