At y = 1 cm, y = 3 cm y = 9 cm, y = 27 cm … and so on , an infinite number of charges equal to 5C are placed. At x = 1 cm, x = 2 cm, x = 4 cm, x = 8 cm …. And so on, an infinite number of charges equal to - 5C are placed. Find the electirc potential at origin is volts . `(K = (1)/(4pi epsilon_(0)))`

Options

A. 250 K
B. `-250 K`
C. zero
D. 100 K

Answer

To find the electric potential at the origin due to the infinite number of charges placed along the y-axis and x-axis, we can follow these steps:

### Step 1: Identify the Charges and Their Positions
- On the y-axis, charges of +5 C are placed at positions \( y = 1 \, \text{cm}, 3 \, \text{cm}, 9 \, \text{cm}, 27 \, \text{cm}, \ldots \) (which can be expressed as \( y = 3^n \) for \( n = 0, 1, 2, \ldots \)).
- On the x-axis, charges of -5 C are placed at positions \( x = 1 \, \text{cm}, 2 \, \text{cm}, 4 \, \text{cm}, 8 \, \text{cm}, \ldots \) (which can be expressed as \( x = 2^n \) for \( n = 0, 1, 2, \ldots \)).

### Step 2: Calculate the Electric Potential from the Charges on the x-axis
The electric potential \( V_x \) at the origin due to the charges on the x-axis is given by the formula:
\[ V_x = k \sum_{n=0}^{\infty} \frac{-5}{r_n} \]
where \( r_n = 2^n \) (the distance from the origin to each charge on the x-axis).

Thus,
\[ V_x = k \left(-5\right) \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \right) \]

The series \( \sum_{n=0}^{\infty} \frac{1}{2^n} \) converges to 2:
\[ V_x = -10k \]

### Step 3: Calculate the Electric Potential from the Charges on the y-axis
The electric potential \( V_y \) at the origin due to the charges on the y-axis is given by:
\[ V_y = k \sum_{n=0}^{\infty} \frac{5}{r_n} \]
where \( r_n = 3^n \) (the distance from the origin to each charge on the y-axis).

Thus,
\[ V_y = k \left(5\right) \left( \frac{1}{1} + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \right) \]

The series \( \sum_{n=0}^{\infty} \frac{1}{3^n} \) converges to \( \frac{3}{2} \):
\[ V_y = 5k \cdot \frac{3}{2} = \frac{15k}{2} \]

### Step 4: Calculate the Total Electric Potential at the Origin
The total electric potential \( V \) at the origin is the sum of the potentials from both axes:
\[ V = V_x + V_y = -10k + \frac{15k}{2} \]

### Step 5: Simplify the Expression
To combine the terms, we can express \( -10k \) as \( -\frac{20k}{2} \):
\[ V = -\frac{20k}{2} + \frac{15k}{2} = -\frac{5k}{2} \]

### Final Answer
Thus, the electric potential at the origin is:
\[ V = -\frac{5k}{2} \text{ volts} \]