Physics, 26.02.2020 23:16 NotCJ126

A +20-nC point charge is placed on the x axis at x = 2.0 m, and a –25-nC point charge is placed on the y axis at y = –3.0 m. What is the direction of the electric field at the origin?

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Answer from: athenadailey8717

Explanation:

Formula for the field at the origin due to the point charge at x-axis is represented as follows.

$E_{x} = \frac{1}{4\pi \epsilon_{o}}(\frac{q_{1}}{r^{2}}$

Putting the given values into the above formula as follows.

$E_{x} = \frac{1}{4\pi \epsilon_{o}}(\frac{q_{1}}{r^{2}}$

= $\frac{1}{4 \times 3.14 \times 8.85 \times 10^{-12}}(\frac{20 \times 10^{-9}}{(2)^{2}}$

= 44.981 N/C

This field is towards negative x-axis as field from positive charge are in a direction outwards from them. Hence, expression for this electric field will be written as follows.

$\vec{E}_{1} = -\vec{E_{x}}\hat{i}$

= $-44.981\hat{i}$

Now, field present due to the charge present at negative y-axis field at the origin is as follows.

$E_{-y} = \frac{1}{4 \pi \epsilon_{o}} \times \frac{25 \times 10^{-9}}{9}$

= $\frac{25 \times 1000}{1000.404}$

= 24.989 N

From negative charge, the field towards negative y-axis is as follows.

$\hat{E_{2}} = -E_{-y}\hat{j}$

= $-24.989\hat{j}$

Now, the net field at the origin is as follows.

$\vec{E_{n}} = \vec{E_{1}} + \vec{E_{2}}$

= $-44.981\hat{i} - 24.989\hat{j}$

Hence, the direction of net field is w.r.t to x-axis is as follows.

$\theta = tan^{-1}(\frac{E_{x}}{E_{y}})$

= $tan^{-1}(\frac{24.989}{44.981})$

= $29.054^{o}$ w.r.t negative x-axis

Therefore, we can conclude that the direction of the electric field at the origin is negative x-axis.

Answer from: Quest

Ithink the answer would be a

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A +20-nC point charge is placed on the x axis at x = 2.0 m, and a –25-nC point charge is placed on t...

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