Answer:
Explanation:
Capacitance of capacitor
= ε₀ A / d , ε₀ is permittivity of space , A is area of plate , d is distance between plates.
= 8.85 x 10⁻¹² x 2 x 10⁻³ / (6 x 10⁻³)
= 2.95 x 10⁻¹² F
Electric field E = V / d , V is potential difference
V = E x d
= 100 x 10³ x 6 x 10⁻³
= 600 V
Charge on the capacitor
= capacitance x potential difference
= 2.95 x 10⁻¹² x 600
= 17.7 x 10¹⁰ C
<h2>Electrostatic Potential Decreases</h2>
Explanation:
- If the spacing between two closely spaced oppositely charged parallel plates is decreased the electrostatic potential difference between the plates will decrease.
- An electrostatic potential that is also referred to as the electric field potential or potential drop is the amount of work required to replace a unit of charge from a reference point to a specific point inside the electric field without any change in acceleration.
- Therefore, if the distance will decrease between oppositely charged plates there will be more affinity to attract which will reduce the amount of work done thus decreasing the electric potential
∴ The Correct option is (b)
D)
<span>No. the products include He-4, one neutron, and energy.</span>
Yes because if 0*c equals 32*F than the higher the number the hotter it is
Answer:
Hi myself Shrushtee.
Explanation:
The fuse is connected to the live wire so that the appliance will not become charged (have a potential difference of 230 V) after the fuse has melted due to excessive current. Fuses must be fitted onto the live wire so that when it blows, it will disconnect (isolate) the appliance from the high voltage live wire.
