Physics: Four +9 µC point charges are at the corners of a square of side 4 m. Find the potential at the center of the square (relative to zero potential at infinity) for each of the following conditions.(a) All the charges are positive (b) Three of the charges are positive and one is negative (c) Two are positive and two are negative

Four +9 µC point charges are at the corners of a

Ответ:

bhenretty2458

23.01.2022

Physics

Answer with Explanation:

We are given that

$q_1=q_2=q_3=q_4=q=9\mu C=9\times 10^{-6} C$

$1\mu C=10^{-6} C$

Side of square,d=4 m

Distance of charge from the center of square,d= $\frac{1}{2}\sqrt{4^2+4^2}=2\sqrt 2m$

Potential is a scalar quantity therefore, we consider sign .

a.When all charges are positive

Then, Potential at center of square,V= $4\times \frac{kq}{d}$

Where $k=9\times 10^9$

$V=4\times \frac{9\times 10^9\times 9\times 10^{-6}}{2\sqrt 2}=1.15\times 10^5 V$

b.Potential at center of square,V= $3\frac{kq}{d}-\frac{kq}{d}=\frac{2kq}{d}$

$V=\frac{2\times 9\times 10^9\times 9\times 10^{-6}}{2\sqrt 2}=5.7\times 10^{4} V$

c.Potential at center of square,V= $\frac{2kq}{d}-\frac{2kq}{d}=0$

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Ответ:

ethanyayger

16.01.2023

Physics

${ \tt{initial \: velocity, \: u = 0}} \: (at \: rest) \\ { \tt{final \: velocity, \: v = 15 { {ms}^{ - 1} }}} \\ { \tt{time, \: t = 6s}} \\ { \bf{from \: first \: newtons \: equation \: of \: motion : }} \\ { \bf{v = u + at}} \\ { \tt{15 = 0 + (a \times 6)}} \\ { \tt{6a = 15}} \\ { \tt{acceleration, \: a = 2.5 \: {ms}^{ - 2} }}$

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