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balu736 [363]
3 years ago
7

Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu

m, each string makes and angle θ =50 with the vertical. Find the size of the charge on each spere.

Physics
1 answer:
Citrus2011 [14]3 years ago
8 0

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

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Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th
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Answer:

The magnitude of the electric field are 2.38\times10^{4}\ N/C and 1.09\times10^{4}\ N/C

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell q_{inn}=3.50\times10^{-8}\ C

Charge on outer shell q_{out}=1.60\times10^{-8}\ C

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

E=\dfrac{kq}{r^2}

Where, q = charge

k = constant

r = distance

Put the value into the formula

E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}

E=2.38\times10^{4}\ N/C

The total charge enclosed  by a radial distance 20.5 cm

The total charge is

q=q_{inn}+q_{out}

Put the value into the formula

q=3.50\times10^{-8}+1.60\times10^{-8}

q=5.1\times10^{-8}\ C

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

E=\dfrac{kq}{r^2}

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E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}

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Answer:

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Explanation:

We are going to use superposition

So, we will first open circuit the current source and find the voltage Voc.

So, check attachment for open circuit diagram.

From the diagram

We notice that R3 is in series with R4, so its equivalent is given below

Req(3-4) = R3 + R4

R(34) = 20+40 = 60 kΩ

Notice that R2 is parallel to the equivalent of R3 and R4, then, the equivalent of all this three resistor is

Req(2-3-4) = R2•R(34)/(R2+R(34))

R(234) = (100×60)/(100+60)

R(234) = 37.5 kΩ

We notice that R1 and R(234) are in series, then, we can apply voltage divider rule to find voltage in R(234)

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V(234) = R(234) / [R1 + R(234)] × V

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V(234) = 37.5/62.5 × 100

V(234) = 60V.

Note, this is the voltage in resistor R2, R3 and R4.

Note that, R2 is parallel to R3 and R4. Parallel resistor have the same voltage, then voltage across R2 equals voltage across R34

V(34) = 60V.

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So we can know the voltage across R4 which is the Voc we are looking for.

Using voltage divider

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Now, finding the short circuit voltage when we short circuit the voltage source

Check attachment for circuit diagram.

From the circuit we notice that R1 and R2 are in parallel, so it's equivalent becomes

Req(1-2) = R1•R2/(R1+R2)

R(12) = 25×100/(25+100)

R(12) = 20 kΩ

We also notice that the equivalent of Resistor R1 and R2 is in series to R3. Then, the equivalent resistance of the three resistor is

Req(1-2-3) = R(12) + R(3)

R(123) = 20 + 20

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We notice that, the equivalent resistance of the resistor R1, R2, and R3 is in series to resistor R4.

So using current divider rule to find the current in resistor R4.

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Then, using ohms law, we can find the voltage across the resistor 4 and the voltage is the required Voc

V = IR

V4 = Voc = I4 × R4

Voc = 4×10^-3 × 40×10^3

Voc = 160V

Then, the sum of the short circuit voltage and the open circuit voltage will give the required Voc

Voc = Voc(open circuit) + Voc(short circuit)

Voc = 24 + 160

Voc = 184V.

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