An electric field `vec E=20y hat j` exist in space . The potential at (5 m, 5 m) is taken to be zero. The potential at origin is

Options

A. 500 V
B. 250 V
C. 125 V
D. Zero

Answer

To find the potential at the origin given the electric field \( \vec{E} = 20y \hat{j} \) and the potential at the point (5 m, 5 m) is zero, we can follow these steps:

### Step 1: Understand the Electric Field
The electric field \( \vec{E} = 20y \hat{j} \) indicates that the electric field is in the positive y-direction and its magnitude depends on the y-coordinate.

### Step 2: Identify the Points
We know that the potential at point \( P(5, 5) \) is zero. We need to find the potential at the origin \( O(0, 0) \).

### Step 3: Use the Relation Between Electric Field and Potential
The relationship between electric field and potential is given by:
\[ dV = -\vec{E} \cdot d\vec{r} \]
In our case, since the electric field is only in the y-direction, we can express this as:
\[ dV = -E_y \, dy \]
where \( E_y = 20y \).

### Step 4: Set Up the Integral
To find the potential difference \( V_O - V_P \) (where \( V_O \) is the potential at the origin and \( V_P \) is the potential at point P), we can integrate from \( y = 5 \) to \( y = 0 \):
\[ V_O - V_P = -\int_{5}^{0} E_y \, dy = -\int_{5}^{0} 20y \, dy \]

### Step 5: Calculate the Integral
Calculating the integral:
\[ V_O - V_P = -\left[ 20 \cdot \frac{y^2}{2} \right]_{5}^{0} = -\left[ 10y^2 \right]_{5}^{0} \]
Now, substituting the limits:
\[ = -\left( 10(0^2) - 10(5^2) \right) = -\left( 0 - 250 \right) = 250 \]

### Step 6: Find the Potential at the Origin
Since \( V_P = 0 \) (the potential at point P), we have:
\[ V_O = V_P + 250 = 0 + 250 = 250 \, \text{V} \]

Thus, the potential at the origin is **250 V**.

### Summary
The potential at the origin is \( V = 250 \, \text{V} \).

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