Question

A −15 nC charge is at x=+2.0 cm on the x-axis. A second charge q is located somewhere on the x-axis to the left of the origin. The electric field at y=2.0 cm on the y-axis is E→=3.0×105i^\overrightarrow{E}=3.0\$$times10^5$$\hat{i}N/C . What are (a) the charge q in nC and (b) its distance from the origin?

Accepted Answer

Step 1: Understand the problem setup. The electric field at a point on the y-axis is influenced by two charges: one at x = +2.0 cm (−15 nC) and another charge q located somewhere on the x-axis to the left of the origin. The electric field at y = 2.0 cm is given as Ē = 3.0×10⁵ î N/C, which means the field is entirely in the x-direction. Step 2: Write the expression for the electric field due to a point charge. The electric field due to a charge q at a distance r is given by: E = k * |q| / r², where k is Coulomb's constant (k = 8.99 × 10⁹ N·m²/C²). The direction of the field depends on the sign of the charge. Step 3: Break down the contributions to the electric field at the point (0, 2.0 cm). The charge at x = +2.0 cm creates an electric field at y = 2.0 cm. Use the distance formula to calculate the distance from the charge to the point: r = √((x₂ - x₁)² + (y₂ - y₁)²). For the charge at x = +2.0 cm, the distance is r = √((2.0 cm)² + (2.0 cm)²). Similarly, the unknown charge q contributes to the field, and its distance from the point (0, 2.0 cm) must also be calculated. Step 4: Use the principle of superposition to combine the electric fields. The total electric field at y = 2.0 cm is the vector sum of the fields due to both charges. Since the field is entirely in the x-direction, the y-components of the fields must cancel out. This allows you to set up equations to solve for the magnitude and position of q. For the x-components, use Eₓ = E₁ₓ + E₂ₓ, where E₁ₓ is the field due to the −15 nC charge and E₂ₓ is the field due to the unknown charge q. Step 5: Solve the equations. Substitute the known values (charge of −15 nC, given electric field, and distances) into the equations. Use algebra to solve for the magnitude of q and its position on the x-axis. Remember to convert units (e.g., cm to m) where necessary and ensure consistency in calculations.

A −15 nC charge is at x=+2.0 cm on the x-axis. A second charge q is located somewhere on the x-axis to the left of the origin. The electric field at y=2.0 cm on the y-axis is $\vec{E} = 3.0 \times 10^{5} \hat{i}$ N/C . What are (a) the charge q in nC and (b) its distance from the origin?

Risposta video verificata per un problema simile:

Concetti chiave

Electric Field

Coulomb's Law

Superposition Principle