Given the electrostatic potential V = 2x2y3 + 3xy2, a) find the electrostatic field strength E, b) calculate the potential at (1,2), and c) calculate the electric field strength at (1,2).

The electric field strength $\vec{E}$ is the negative gradient of the electric potential $V$:

The electrostatic field strength is $\boxed{\vec{E} = -(4xy^3 + 3y^2)\hat{i} - (6x^2y^2 + 6xy)\hat{j}}$.

b) Calculate the electric potential at the position $x=1, y=2$.

The electric potential at $(1,2)$ is $\boxed{\text{28 V}}$.

c) Calculate the electrostatic field strength in the same position, $x=1, y=2$.

The electrostatic field strength at $(1,2)$ is $\boxed{\vec{E} = -44\hat{i} - 36\hat{j} \, \text{N/C}}$.

d) If a negative charge $q = -2.3 \times 10^{-5} \, \text{C}$ is located at $x=1, y=2$, calculate the electrostatic force on the charge.

Given charge $q = -2.3 \times 10^{-5} \, \text{C}$.
From part (c), $\vec{E} = -44\hat{i} - 36\hat{j} \, \text{N/C}$ at $(1,2)$.

The electrostatic force on the charge is $\boxed{\vec{F} = (1.012 \times 10^{-3})\hat{i} + (0.828 \times 10^{-3})\hat{j} \, \text{N}}$.

28.

a) What is corona discharge?
Corona discharge is an electrical discharge caused by the ionization of a fluid (such as air) surrounding a conductor, which occurs when the electric field strength is high enough to cause ionization but not high enough to cause a complete electrical breakdown or arc. It often appears as a faint glow.

b) Explain how corona discharge is applied in the making of a Van de Graaff electrostatic generator.
In a Van de Graaff generator, corona discharge is used at two points:

• At the lower comb, sharp points create a very high electric field, causing air molecules to ionize. This allows charge to be sprayed from the comb onto the moving insulating belt.
• At the upper comb, another set of sharp points creates a high electric field, causing corona discharge that collects the charge from the belt and transfers it to the large metallic sphere, building up a high potential.

c) i) Take two metal spheres A and B of radius r and R respectively where R is bigger than r (R > r). Which of A or B will take smaller quantity of charge to be raised to a given potential and why?
Sphere A (with radius $r$) will take a smaller quantity of charge to be raised to a given potential.
The potential $V$ of an isolated conducting sphere with charge $Q$ and radius $r$ is given by $V = \frac{kQ}{r}$, where $k$ is Coulomb's constant. Rearranging for charge, $Q = \frac{Vr}{k}$. For a given potential $V$, the charge $Q$ is directly proportional to the radius $r$. Since sphere A has a smaller radius ($r < R$), it requires less charge to reach the same potential compared to sphere B.

ii) Calculate the charge required to raise the potential of an isolated metallic sphere of radius 2m from zero to $3.4 \times 10^5 \, \text{V}$.

Radius ($r$) = $2 \, \text{m}$.
Potential ($V$) = $3.4 \times 10^5 \, \text{V}$.
Coulomb's constant ($k$) = $9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$.
The potential of an isolated sphere is $V = \frac{kQ}{r}$.

The charge required is $\boxed{\text{7.56} \times 10^{-5} \, \text{C}}$.