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3 years ago

If you placed a negatively charged object within this electric field, which direction will it move?

Physics

2 answers:

kramer3 years ago

6 0

Option A would be the right answer.

Alex787 [66]3 years ago

4 0

Answer: Option (A) to the right

Explanation:

Let me explain it in simple words!

Always remember that the electric field direction is outward from a positive charge (be it a single positively charged particle, or a positive rod) and inward into a negative charge. In this case, as you can see, the electric field arrows' direction is from right to left. It means that the positive charge is on the right side, and the negative charge is on the left side.

Now what will happen when you place a negatively charged object within this electric field? Well, as you know, two negatively charged objects repel each other, and positively and negatively charged objects attract each other. The negatively charged object will move<em> towards right</em>, since there is a positive charge on right side (as explained in the first paragraph), which will attract this negatively charged object. Furthermore, the negative charge on the left side will repel this negatively charged object towards right (against the direction of the electric field). Hence, the correct answer is Option (A) to the right.

You must be wondering why I didn't mention the upward and downward motion. Well, again always remember that the positively charged object moves along the direction of electric field, and the negatively charged object moves against the direction of electric field (as explained above). <em>Charged bodies do not move perpendicular to the direction of electric field</em> (unless there is some net external force applied to that object).

I hope this helps! :)

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Frictional force, F = 45.9 N

Explanation:

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The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

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Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

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Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

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$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

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$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

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