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2 years ago

What is the formula for a uniform electric field between two parallel plates?

Physics

1 answer:

vlabodo [156]2 years ago

6 0

Answer: B. V/b
Explanation:

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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

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Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

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2 years ago

17 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What

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Answer:

$\frac{D}{d} = 4.12$

Explanation:

As we know that resistance of one copper wire is given as

$r = \rho \frac{L}{a}$

here we know that

$a = \pi (\frac{d}{2})^2$

now we have

$r = \rho \frac{L}{\pi (\frac{d^2}{4})}$

$r = \rho \frac{4L}{\pi d^2}$

now we know that such 17 resistors are connected in parallel so we have

$R = \frac{r}{17}$

$R = \rho \frac{4L}{17 \pi d^2}$

Now if a single copper wire has same resistance then its diameter is D and it is given as

$R = \rho \frac{4L}{\pi D^2}$

now from above two equations we have

$\rho \frac{4L}{\pi D^2} = \rho \frac{4L}{17 \pi d^2}$

$D^2 = 17 d^2$

now we have

$\frac{D}{d} = 4.12$

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