Question

The electric potential varies in space according to the relation $\mathrm{V}=3 \mathrm{x}+4 \mathrm{y}$. A particle of mass 10 kg starts from rest from point (2,3.2) $\mathrm{m}$ under the influence of this field. The velocity of the particle when it crosses the x - axis is equal to $0 x \times 10^{-3} \mathrm{m} / \mathrm{s}$. The charge on the particle is $+1 \mu \mathrm{C}$ Assume $\mathrm{V}(\mathrm{x}, \mathrm{y})$ are in $\mathrm{Sl}$ units. Find the value of x.

Answer 1

We know that in physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. On the other hand, potential energy is energy that is stored or conserved in an object or substance. This stored energy is based on the position, arrangement or state of the object or substance. We can think of it as energy that has the 'potential' to do work. Based on this concept we have to solve this question.

Complete step by step answer
It is given that:
$V=3 x+4 y$
So, the initial position is $(2,3.2),$
Hence, we can say that, $V_{1}=3 \times 2+4 \times 3.2=18.8 \mathrm{V}$
Now, $\mathrm{E}_{\mathrm{x}}=\dfrac{-\partial \mathrm{V}}{\partial \mathrm{x}}=-3 \mathrm{V}$
and $\mathrm{E}_{\mathrm{y}}=\dfrac{-\partial \mathrm{V}}{\partial \mathrm{y}}=-4 \mathrm{V}$
Hence, the electric field is given as: $\overrightarrow{\mathrm{E}}=-3 \hat{\mathrm{i}}-4 \hat{\mathrm{i}}$
Hence, the path of charge will be along this vector as shown in figure.
Now, equation of line of travel of charge is given as:
The slope is given as $=\dfrac{4}{3}$
$y-3.2=\dfrac{4}{3}(x-2)$
When, point passes through $\mathrm{X}$ -axis, $\mathrm{y}=0:-$
Therefore, the equation becomes:
$\Rightarrow -3.2=\dfrac{4}{3}(\text{x}-2)$
$\Rightarrow \text{x}=-0.4$
Here, potential=V $_{2}=3 \times(-0.4)+4 \times 0=-1.2 \mathrm{V}$
Now, gain in kinetic energy = loss in potential energy
$\Rightarrow {{\text{K}}_{2}}-{{\text{K}}_{1}}={{\text{V}}_{1}}-{{\text{V}}_{2}}$
So, after we put the values, we get that:
$\Rightarrow \dfrac{1}{2}\text{m}{{\text{v}}^{2}}=\text{q}\\{18.8-(-1.2)\\}=20\text{q}$
Now let us evaluate the above expression:
$\Rightarrow {{\text{v}}^{2}}=\dfrac{40\times {{10}^{-6}}}{10}$
$\Rightarrow {{\text{v}}^{2}}=4\times {{10}^{-6}}$
$\Rightarrow \text{v}=2\times {{10}^{-3}}\text{m}/\text{s}$

Note: We should know that kinetic energy is built up in an object by motion and can be defined as the energy that is needed to either slow it down or speed it up. Studies have also been done using kinetic energy and then converting it to other types of energy, which is then used to power everything from flashlights to radios and more. Energy is transferred from one object to another when a reaction takes place. Energy comes in many forms and can be transferred from one object to another as heat, light, or motion, to name a few.
On the other hand, potential energy becomes useful when it is converted to kinetic energy. Kinetic energy is used to perform work. Work occurs when force is applied to an object. Kinetic energy is useful because it allows work to be done. Potential energy is the energy by virtue of an object's position relative to other objects. Potential energy is often associated with restoring forces such as a spring or the force of gravity. This work is stored in the force field, which is said to be stored as potential energy.