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Nataliya [291]
3 years ago
11

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

ach sphere if the threads make an angle θ of 15 degrees with the vertical?
Physics
1 answer:
Nataliya [291]3 years ago
5 0

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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Answer: 13.6 N

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a_{x}=\frac{2 \Delta x}{t^{2}} (2)

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A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the
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Answer:

(a). The electric potential at 1.650 cm is -1.219\times10^{-4}\ V.

(b). The electric potential at 2.81 cm is -3.759\times10^{-4}\ V.

Explanation:

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(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr

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V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})

V_{r}=-0.00012190\ V

V_{r}=-1.219\times10^{-4}\ V

The electric potential at 1.650 cm is -1.219\times10^{-4}\ V.

(b). We need to calculate the potential at a distance r = R

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V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}

V_{R}=-0.0003759\ V

V_{R}=-3.759\times10^{-4}\ V

The electric potential at 2.81 cm is -3.759\times10^{-4}\ V.

Hence, This is the required solution.

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