Question

TWO EQUAL POSITIVE CHARGES ARE KEPT AT POINTS A AND B. THE ELECTRIC POTENTIAL AT THE POINTS BETWEEN A AND B IS STUDIED WHILE MOVING FROM A TO B. THE POTENTIAL-
1. CONTINUOUSLY INCREASES
2. CONTINUOUSLY DECREASES
3. INCREASES THEN DECREASES
4. DECREASES THEN INCREASES

Answer 1

Answer on Question#40102, Physics, Electrodynamics

TWO EQUAL POSITIVE CHARGES ARE KEPT AT POINTS A AND B. THE ELECTRIC POTENTIAL AT THE POINTS BETWEEN A AND B IS STUDIED WHILE MOVING FROM A TO B. THE POTENTIAL-

1. CONTINUOUSLY INCREASES

2. CONTINUOUSLY DECREASES

3. INCREASES THEN DECREASES

4. DECREASES THEN INCREASES

Solution

Let's consider point $x$ between the charges. The distance between the first charge (A) and point $X$ is $x$. The distance between the second charge (B) and point $X$ is $r - x$.

The electric potential at any point is the algebraic sum of the potential at that point due to each individual charge. Therefore, the potential at the point $x$ between the charges is

$$V = \frac {k q}{x} + \frac {k q}{r - x}.$$

Let's find derivative of $V(x)$:

$$\frac {d V}{d x} = \frac {d}{d x} \left(\frac {k q}{x} + \frac {k q}{r - x}\right) = k q \left(- \frac {1}{x ^ {2}} + \frac {1}{(r - x) ^ {2}}\right) = k q \frac {x ^ {2} - (r - x) ^ {2}}{x ^ {2} (r - x) ^ {2}}.$$

This function is negative when $x < \frac{r}{2}$ - the potential decreases to middle of distance, the function is positive when $x > \frac{r}{2}$ - the potential increases to point B.

Answer: 4. DECREASES THEN INCREASES.