remains the same
Step 1: The electrostatic potential energy ($U$) between two point charges $q_1$ and $q_2$ separated by distance $r$ is given by: \[ U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} \] Step 2: Since both charges are protons, we have: \[ q_1 = q_2 = e = 1.6 \times 10^{-19} { C} \]
Step 3: Since both charges are positive, they repel each other. As the distance between them decreases, $r$ decreases.
Step 4: Since $U \propto \frac{1}{r}$, decreasing $r$ increases $U$, meaning the electrostatic potential energy of the system increases.
Step 5: Therefore, the correct answer is (B). \bigskip
\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]
Then \( f'(-4) \) is equal to:If \( f'(x) = 4x\cos^2(x) \sin\left(\frac{x}{4}\right) \), then \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
\[ \int \frac{4x \cos \left( \sqrt{4x^2 + 7} \right)}{\sqrt{4x^2 + 7}} \, dx \]