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laiz [17]
2 years ago
15

Derive the value of electric field due to a uniform sphere of charge.

Physics
1 answer:
Deffense [45]2 years ago
4 0
<h2>Hey there!</h2>

The Force "F" applied on the unit electric charge "q" at a point describes the electric field.

<h3>☆ Formula to find electric charge:</h3>

  • E = F/q

<h2>Hope it helps </h2>
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The tension in the horizontal cord is 30N. Find the weight of the object.
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Making a drawing of the system, we will have two forces which are tension and the weight of the object. Balancing the forces present, we do as follows:

T = W
W = 30 N

Therefore, weight is equal to 30 N. Hope this answers the question. Have a nice day. Feel free to ask more questions.
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At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
lbvjy [14]

Answer:

F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

|F_T|=2\sqrt{34}k\frac{Q^2}{L}

\theta=tan^{-1}(\frac{5}{3})=59.03\°

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

and the magnitude is:

|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}

the direction is:

\theta=tan^{-1}(\frac{5}{3})=59.03\°

4 0
3 years ago
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