Physics, 06.12.2019 01:31 simmy6

Two point charges are located along the x axis: q1 = +5.9 μc at x1 = +4.2 cm, and q2 = +5.9 μc at x2 = −4.2 cm. two other charges are located on the y axis: q3 = +2.9 μc at y3 = +4.8 cm, and q4 = −8.2 μc at y4 = +6.9 cm. find the net electric field (magnitude and direction) at the origin.

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Answer from: morganashley6795

Explanation:

In order to find the net electric field, we have to find the four electric fields and add them as vecotrs.

Hence, formula to calculate the electric field is as follows.

E = $\frac{kq}{r^{2}}$

For the charge at x = 0.042 m (as 1 m = 100 cm)

So, $E_{1} = \frac{9 \times 10^{9} \times 5.9}{(0.042)^{2}}$

= $3.01 \times 10^{12}$ N (toward the negative x direction)

For the charge at x = -0.042 m

$E_{2} = \frac{9 \times 10^{9} \times 5.9}{(0.042)^{2}}$

= $3.01 \times 10^{12}$ N (toward the positive x direction)

For the charge at y = 0.048 m

$E_{3} = \frac{9 \times 10^{9} \times 2.9}{0.048}^{2}}$

= $11.32 \times 10^{12}$ N (toward the negative y direction)

For the charge at y = 0.069 m

$E_{4} = \frac{9 \times 10^{9} \times 8.2}{0.069)^{2}}$

= $15.50 \times 10^{12}$ N (toward the positive y)

Therefore, net x is calculated as follows.

The net x = $3.01 \times 10^{12} N - 3.01 \times 10^{12} N$

= 0 N

The net y = $15.50 \times 10^{12} N - 11.32 \times 10^{12}$ N

= $4.18 \times 10^{12}$ N (positive y)

Total net E is found by the pythagorean theorem as follows.

E = $\sqrt{[(0 N)^{2} + (4.18 \times 10^{12} N)^{2}]$

= $4.18 \times 10^{6}$ N

Thus, we can conclude that the net electric field (magnitude and direction) at the origin is $4.18 \times 10^{6}$ N.

Two point charges are located along the x axis: q1 = +5.9 μc at x1 = +4.2 cm, and q2 = +5.9 μc at x

Answer from: Hfruit

Explanation:

Given

$q_1=+5.9\mu C$

$q_2=+5.9\mu C$

$q_3=+2.9\mu C$

$q_4=-8.2\mu C$

Electric field is given by

$E=\frac{kq}{r^2}$

Electric field due to q_1 and q_2 cancel out each other because they point in opposite direction in x-axis

so net Electric field will be because of q_3 and q_4 towards negative Y axis

$E_3=\frac{kq_3}{r_3^2}$

$E_3=\frac{9\times 10^9\times 2.9\times 10^{-6}}{(4.8\times 10^{-2})^2}$

$E_3=1.132\times 10^7$ (away from q_3[/tex])

similarly $E_4=1.55\times 10^{7}$ (towards q_4[/tex])

$E_{net}=E_4-E_3=1.55\times 10^{7}-1.132\times 10^7$

$E_{net}=0.41\times 10^7 N/C$

Answer from: Quest

what experimental evidence led scientists to change from the previous model to this one?

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Answer from: Quest

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Two point charges are located along the x axis: q1 = +5.9 μc at x1 = +4.2 cm, and q2 = +5.9 μc at x...

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