To find the potential at point B given the potential at point A is zero in the electric field \( \vec{E} = 10x \hat{i} - 20 \hat{j} \, \text{N/C} \), we can use the relationship between electric potential and electric field. The potential difference between two points A and B is given by:
\[
V_B - V_A = -\int_A^B \vec{E} \cdot d\vec{r}
\]
### Step 1: Identify the coordinates
- Point A: \( (30 \, \text{m}, -20 \, \text{m}) \)
- Point B: \( (-20 \, \text{m}, 30 \, \text{m}) \)
### Step 2: Define the displacement vector
The displacement vector \( d\vec{r} \) can be expressed as:
\[
d\vec{r} = dx \hat{i} + dy \hat{j}
\]
### Step 3: Calculate the dot product \( \vec{E} \cdot d\vec{r} \)
The electric field vector is:
\[
\vec{E} = 10x \hat{i} - 20 \hat{j}
\]
Now, we compute the dot product:
\[
\vec{E} \cdot d\vec{r} = (10x \hat{i} - 20 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = 10x \, dx - 20 \, dy
\]
### Step 4: Set up the integral
The potential difference can be expressed as:
\[
V_B - V_A = -\int_A^B (10x \, dx - 20 \, dy)
\]
Since \( V_A = 0 \), we have:
\[
V_B = -\int_A^B (10x \, dx - 20 \, dy)
\]
### Step 5: Determine the limits of integration
- For \( x \): from \( x_A = 30 \) to \( x_B = -20 \)
- For \( y \): from \( y_A = -20 \) to \( y_B = 30 \)
### Step 6: Evaluate the integral
We can separate the integral:
\[
V_B = -\left( \int_{30}^{-20} 10x \, dx + \int_{-20}^{30} (-20) \, dy \right)
\]
Calculating the first integral:
\[
\int 10x \, dx = 5x^2
\]
Evaluating from 30 to -20:
\[
5[-20^2 - 30^2] = 5[-400 - 900] = 5[-1300] = -6500
\]
Calculating the second integral:
\[
\int -20 \, dy = -20y
\]
Evaluating from -20 to 30:
\[
-20[30 - (-20)] = -20[50] = -1000
\]
### Step 7: Combine the results
Now, substituting back into the equation for \( V_B \):
\[
V_B = -(-6500 - 1000) = 7500 \, \text{V}
\]
### Final Answer
The potential at point B is:
\[
V_B = 7500 \, \text{V}
\]
