Question

A point 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

Answer 1

Answer:

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, $E$ is the electric field, $q$ is the magnitude of charge and $r$ 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 $\boxed{562.5\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}}{\text{C}}}} \right.\kern-\nulldelimiterspace} {\text{C}}}}$ .

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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 2

The net electric field between the two point charges is 562.5 N/C.

What is an Electric Field?

The region where an electrostatic force of attraction or repulsion is experienced on the particles is known as an electric field.

Given data:

The magnitude of the first point charge is, q = - 4.00 nC.

The distance for the first point charge is, x = 0.600 m.

The magnitude of the second point charge is, q' = +6.00 nC.

The distance along the y-axis is, y = 0.800 m.

The expression for the net electric field between the two point charges is given as,

$E= E_{1}+E_{2}\\\\E = \dfrac{kq}{x^{2}}+ \dfrac{kq'}{x^{2}}$

Here,

k is the Coulomb's constant.

Solving as,

$E = \dfrac{9 \times 10^{9}}{0.600^{2}} \times (-4.00 + 6.00) \times 10^{-9}\\\\E = 562.5 \;\rm N/C$

Thus, we can conclude that the net electric field between the two point charges is 562.5 N/C.

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