Physics, 08.07.2019 11:30 nique0808

Apoint charge q1=−4.00nc is at the point x=0.600 meters, y=0.800 meters, and a second point charge q2=+6.00nc is at the point x=0.600 meters, y=0. calculate the magnitude e of the eectric field between these two charges

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Answer from: hwhite41

The net electric field at the mid-point of the two charges is $\boxed{562.5\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}} {\text{C}}}} \right.\kern-\nulldelimiterspace} {\text{C}}}}$ .

Further Explanation:

Given:

The magnitude of the charge ${q_1}$ is $- 4.00\,{\text{nC}}$ .

The magnitude of the charge ${q_2}$ is $+ 6.00\,{\text{nC}}$ .

The co-ordinates of the location of charge ${q_1}$ are $\left( {0.600\,{\text{m}},0.{\text{800}}\,{\text{m}}} \right)$ .

The co-ordinates of the location of charge ${q_2}$ are $\left( {0.600\,{\text{m}},0.0\,{\text{m}}} \right)$ .

Concept:

The electric field due to a positive charge will be in the direction away from it and due to the negative charge, the electric field points in the direction away from the charge. Therefore, the net electric field due to both charges will be in same direction.

Write the expression for the electric field due to a charge.

$\boxed{E = \dfrac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}}}$ …… (1)

Here, is the electric field, is the magnitude of charge and is the distance of the point from the charge.

The distance between the two charges is $0.800\,{\text{m}}$ as shown in the figure. Therefore, the distance of the mid-point from the charges is $0.400\,{\text{m}}$ .

The electric field due to charge ${q_1}$ is:

$\begin{aligned}{E_1} &= \frac{1}{{4\pi {\varepsilon _0}}}\frac{{4 \times {{10}^{ - 9}}}}{{{{\left( {0.400} \right)}^2}}}\\&=\frac{{9 \times {{10}^9} \times 4 \times {{10}^{ - 9}}}}{{0.16}}\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}} {\text{C}}}} \right.\kern-\nulldelimiterspace} {\text{C}}}\\&= 225\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}} {\text{C}}}} \right.\kern-\nulldelimiterspace} {\text{C}}}\\\end{aligned}$

The electric field due to charge ${q_2}$ is:

$\begin{aligned}{E_2} &= \frac{1}{{4\pi {\varepsilon _0}}}\frac{{6 \times {{10}^{ - 9}}}}{{{{\left( {0.400} \right)}^2}}} \\ &=\frac{{9 \times {{10}^9} \times 6 \times {{10}^{ - 9}}}}{{0.16}}\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}} {\text{C}}}} \right.\kern-\nulldelimiterspace} {\text{C}}}\\&=337.5\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}} {\text{C}}}} \right.\kern-\nulldelimiterspace} {\text{C}}}\\\end{aligned}$

The net electric field at the mid-point of the two charges is.

$E = {E_1} + {E_2}$ …… (2)

Substitute the value of ${E_1}$ and ${E_2}$ in above equation.

$\begin{aligned}E&= 225\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}} {\text{C}}}} \right.\kern-\nulldelimiterspace} {\text{C}}} + 337.5\,{{\text{N}}\mathord{\left/{\vphantom {{\text{N}} {\text{C}}}} \right.\kern-\nulldelimiterspace} {\text{C}}}\\&= 562.5\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}} {\text{C}}}} \right.\kern-\nulldelimiterspace} {\text{C}}}\\\end{aligned}$

Thus, the net electric field at the mid-point of the two charges is $\boxed{562.5\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}}{\text{C}}}} \right.\kern-\nulldelimiterspace} {\text{C}}}}$ .

Learn More:

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Answer Details:

Grade:College

Chapter:Electrostatics

Subject:Physics

Keywords:Two charges, electric field, point charges, magnitude of field, E=kq/r^2 , coordinates of location, mid-point of two charges.

Answer from: Quest

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