Given the circuit below. What is the potential difference between point P and the ground?

Let $I_1$ be the current flowing from P through the $10 \, \Omega$ resistor.
Let $I_2$ be the current flowing into the junction from the left ($4 \, \text{A}$).
Let $I_3$ be the current flowing through the $20 \, \Omega$ resistor to the ground.
According to the diagram, the current flowing into point P is $2 \, \text{A}$. This current then flows through the $10 \, \Omega$ resistor. So, $I_1 = 2 \, \text{A}$.
At the junction, by Kirchhoff's Current Law (KCL), the sum of currents entering the junction equals the sum of currents leaving it.
Current entering the junction from the $10 \, \Omega$ resistor is $I_1 = 2 \, \text{A}$.
Current entering the junction from the left is $I_2 = 4 \, \text{A}$.
So, the total current leaving the junction through the $20 \, \Omega$ resistor is $I_3 = I_1 + I_2$.

The $20 \, \Omega$ resistor is connected between the junction J and the ground ($0 \, \text{V}$).
Using Ohm's Law, $V_J = I_3 \times R_{20\Omega}$.

Point P is connected to the junction J through the $10 \, \Omega$ resistor. The current $I_1 = 2 \, \text{A}$ flows from P to J.
So, the potential at P is higher than the potential at J by the voltage drop across the $10 \, \Omega$ resistor.

The correct answer is $\boxed{\text{b) } 140 \, \text{V}}$.

The potential difference from X to Y is $\Delta V = V_Y - V_X = +1.5 \, \text{V} - 0 \, \text{V} = +1.5 \, \text{V}$.

When a charged particle moves in an electric field, the gain in kinetic energy ($\Delta KE$) is equal to the negative of the change in potential energy ($\Delta PE$).

The change in potential energy is given by $\Delta PE = q \Delta V$, where $q$ is the charge of the particle.
So, $\Delta KE = -q \Delta V$.

We are given $\Delta KE = 3.0 \, \text{eV}$ and $\Delta V = 1.5 \, \text{V}$.

To find $q$ in terms of elementary charge $e$, we can divide both sides by $1 \, \text{V}$:

• A proton has a charge of $+e$.
• A positron has a charge of $+e$.
• An electron has a charge of $-e$.
• An alpha particle (helium nucleus) has a charge of $+2e$. None of the standard particles listed have a charge of $-2e$. However, if the question implies a particle that loses $3.0 \, \text{eV}$ of potential energy, then it would gain $3.0 \, \text{eV}$ of kinetic energy. Let's re-evaluate the sign. If a particle gains kinetic energy, its potential energy must decrease. So, $\Delta PE = -3.0 \, \text{eV}$. Then $q \Delta V = -3.0 \, \text{eV}$. $q (+1.5 \, \text{V}) = -3.0 \, \text{eV}$. $q = \frac{-3.0 \, \text{eV}}{1.5 \, \text{V}} = -2e$. This still indicates a particle with charge $-2e$. This is not a common fundamental particle.

Let's consider if the question meant a particle that loses $3.0 \, \text{eV}$ of kinetic energy, or if there's a typo in the options or the question.
If the particle was an electron (charge $-e$), $\Delta KE = -(-e)(1.5 \, \text{V}) = +1.5 \, \text{eV}$.
If the particle was a proton (charge $+e$), $\Delta KE = -(+e)(1.5 \, \text{V}) = -1.5 \, \text{eV}$ (meaning it loses kinetic energy).
If the particle was an alpha particle (charge $+2e$), $\Delta KE = -(+2e)(1.5 \, \text{V}) = -3.0 \, \text{eV}$ (meaning it loses $3.0 \, \text{eV}$ of kinetic energy).

The question states "gains $3.0 \, \text{eV}$ of kinetic energy". This means $\Delta KE = +3.0 \, \text{eV}$.
Our calculation consistently gives $q = -2e$. Since this is not an option, there might be a misunderstanding of the question or a typo.
However, if the question meant "Which particle loses $3.0 \, \text{eV}$ of kinetic energy" or "Which particle gains $3.0 \, \text{eV}$ of potential energy", then an alpha particle would fit.
Given the options, and the commonality of such problems, it's possible the question intended to ask for a particle that loses $3.0 \, \text{eV}$ of kinetic energy when moving from X to Y, or gains $3.0 \, \text{eV}$ of kinetic energy when moving from Y to X.
If it moves from Y to X, $\Delta V = V_X - V_Y = 0 - 1.5 \, \text{V} = -1.5 \, \text{V}$.
Then $\Delta KE = -q \Delta V = -q (-1.5 \, \text{V}) = +q (1.5 \, \text{V})$.
If $\Delta KE = +3.0 \, \text{eV}$, then $q (1.5 \, \text{V}) = 3.0 \, \text{eV}$, which means $q = +2e$.
An alpha particle has a charge of $+2e$. This would make sense if the direction of motion was Y to X.

Assuming the question implies a standard particle and a common scenario, and given the options, the most plausible interpretation is that an alpha particle would gain $3.0 \, \text{eV}$ of kinetic energy if it moved from Y to X. However, the question explicitly states "moving from X to Y".
If we strictly follow "moving from X to Y" and "gains $3.0 \, \text{eV}$ of kinetic energy", then the charge must be $-2e$. Since this is not an option, there is an inconsistency.

Let's consider the possibility that the question meant "Which particle loses $3.0 \, \text{eV}$ of kinetic energy when moving from X to Y?".
If $\Delta KE = -3.0 \, \text{eV}$, then $-3.0 \, \text{eV} = -q (1.5 \, \text{V})$, which means $q = +2e$. This would be an alpha particle.
This is a common type of error in multiple-choice questions. Given the options, an alpha particle is the only one that relates to $3.0 \, \text{eV}$ with a $1.5 \, \text{V}$ potential difference.

The most likely intended answer, assuming a slight misphrasing of the question, is an alpha particle.

The correct answer is $\boxed{\text{d) alpha particle}}$.

Let $Q_1 = +4\mu\text{C}$ and $Q_2 = -16\mu\text{C}$.
The separation distance is $d = 120 \, \text{mm}$.
Let the $+4\mu\text{C}$ charge be at the origin ($x=0$). Then the $-16\mu\text{C}$ charge is at $x=d = 120 \, \text{mm}$.
The electric potential $V$ at a point $x$ due to these two charges is given by:

where $k$ is Coulomb's constant, $r_1$ is the distance from $Q_1$, and $r_2$ is the distance from $Q_2$. We want to find $x$ such that $V(x) = 0$.

In this region, $r_1 = x$ and $r_2 = d-x$.

Substitute the charge values (we can ignore the $\mu\text{C}$ and $k$ as they cancel out):

Cross-multiply:

This point is between the charges and is option (a).

Since the charges have opposite signs and different magnitudes, there will also be a point outside the charges where the potential is zero. This point will be closer to the charge with the smaller magnitude. In this case, $|Q_1| < |Q_2|$, so the point will be to the left of $Q_1$.
Let the distance from $Q_1$ be $x$. So the coordinate of this point is $-x$.
The distance from $Q_1$ is $x$. The distance from $Q_2$ is $d+x$.

Substitute the charge values:

Cross-multiply:

This point is to the left of $Q_1$ and is option (b).

Both $24 \, \text{mm}$ and $40 \, \text{mm}$ are valid distances from the $+4\mu\text{C}$ charge where the electric potential is zero. Since the question asks "How far from the $+4\mu\text{C}$ will the electric potential be zero?" without specifying which point, and both are options, we typically choose the one that is usually considered in introductory problems or the one that is "between" the charges if it exists. In this case, $24 \, \text{mm}$ is between the charges.

The correct answer is $\boxed{\text{a) } 24 \, \text{mm}}$.

Hooke's Law states that the force $F$ applied to a spring or wire is proportional to its extension $x$:

where $k$ is the spring constant (stiffness).
The elastic potential energy $E$ stored in the wire is given by:

From Hooke's Law, $x = \frac{F}{k}$. Substitute this into the energy formula:

When the load is $F$, the initial elastic energy is:

We are given that this initial energy is $E$. So, $E = \frac{F^2}{2k}$.

When the load is increased to $2F$, the final elastic energy is:

Since $E = \frac{F^2}{2k}$, we have:

The increase in elastic energy is $\Delta E = E_2 - E_1$.

The correct answer is $\boxed{3E}$.