Question

a charge of 50 nc is uniformly distributed along the y axis from y = 3.0 m to y = 5.0 m. what is the magnitude of the electric field at the origin?

Answer 1

The magnitude of the electric field at the origin can be calculated by considering the contribution from each segment of the charge distribution along the y-axis. Therefore, the magnitude of the electric field at the origin is approximately 30 N/C. This negative sign indicates that the electric field points in the negative y-direction.

Step-by-step explanation:

To calculate the electric field at the origin due to the uniformly distributed charge along the y-axis, we can use the principle of superposition. The electric field due to a small segment of charge is given by the equation:

`\[ dE = (k \cdot dq)/(r^2) \]`

Where:

• dE = electric field due to a small segment of charge
• k = Coulomb's constant ( 8.9875 × 10⁹ N m²/C²)
• dq = small segment of charge
• r = distance from the small segment of charge to the point where the electric field is being calculated

The total electric field at a point is then found by integrating this expression over the entire distribution of charge.

Given:

- Charge density,
`\( \lambda = 50 nC/m \)`

- Limits of integration,
`\( y_1 = 3.0 m \) and \( y_2 = 5.0 m \)`

We can express dq in terms of dy using the charge density:

`\[ dq = \lambda dy \]`

The distance r from a small segment of charge to the origin is simply ( r = y).

Now, we can integrate to find the total electric field at the origin:

`\[ E = \int_(y_1)^(y_2) dE = k\int_(y_1)^(y_2) (\lambda dy)/(y^2) \]`

Plugging in the given values, we get:

`\[ E = k\lambda\int_(3.0)^(5.0) (dy)/(y^2) \]`

Solving this integral gives us the magnitude of the electric field at the origin.

Calculation:

`\[ E = k\lambda\int_(3.0)^(5.0) (dy)/(y^2) = (8.9875 × 10^9 N m^2/C^2)(50 × 10^(-9) C/m)\left[-(1)/(y)\right]_(3.0)^(5.0) \]`

`\[ E = (8.9875 × 10^9 N m^2/C^2)(50 × 10^(-9) C/m)\left(-(1)/(5.0) + (1)/(3.0)\right) \]`

`\[ E ≈ (8.9875 × 10^9 N m^2/C^2)(50 × 10^(-9) C/m)\left(-(6)/(30)\right) ≈ -30 N/C\]`

So, he magnitude of the electric field at the origin is 30 N/C