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ElenaW [278]
3 years ago
14

Two charged metallic spheres of radii, R = 10 cms and R2 = 20 cms are touching each other. If the charge on each sphere is +100

nC, what is the electric potential energy between the two charged spheres?
Physics
2 answers:
Molodets [167]3 years ago
8 0

Answer:

200\times 10^{-6}j

Explanation:

We have given the radius of first sphere is 10 cm and radius of second sphere is 20 cm

So the potential of first sphere will be greater than the potential of the second sphere, so charge will flow from first sphere to second sphere

Let q charge is flow from first sphere to second sphere and then potential become same

So V=\frac{K(100-q)}{r_1}=\frac{K\times 100}{r_2}

200-100=2q+q

q=\frac{100}{3}=33.33nC

So V=\frac{K(100-q)}{r_1}=\frac{9\times 10^{9}\times (100-33.33)\times 10^{-9}}{10\times 10^{-2}}=6003V

We know that potential energy U=qV=33.33\times 10^{-9}\times 6003=200\times 10^{-6}j

Zolol [24]3 years ago
6 0

Answer:

The electric potential energy between the two charged spheres is 199.9\times10^{-6}\ J

Explanation:

Given that,

Radius of first sphere R_{1}=10\ cm

Radius of second sphere R_{2}=10\ cm

Charge Q= 100 nC

We know charge flows through higher potential to lower potential.

Using formula of potential

V=\dfrac{k(Q-q)}{R_{1}}...(I)

V=\dfrac{k(Q+q)}{R_{2}}...(II)

From equation (I) and (II)

\dfrac{k(Q-q)}{R_{1}}=\dfrac{k(Q+q)}{R_{2}}

Put the value into the formula

\dfrac{(100-q)}{10\times10^{2}}=\dfrac{(100+q)}{20\times10^{-2}}

(100-q)\times20\times10^{-2}=(100+q)\times10\times10^{-2}

q=\dfrac{1000}{30}

q=\dfrac{100}{3}

q=33.33\ nC

So, the potential at R₁ and R₂

Using formula of potential

V=\dfrac{k(Q-q)}{R_{1}}

Put the value into the formula

V=\dfrac{9\times10^{9}(100-33.33)\times10^{-9}}{10\times10^{-2}}

V=6000.3\ Volt

We need to calculate the electric potential energy between the two charged spheres

Using formula of  the electric potential energy

U=qV

U=33.33\times10^{-9}\times6.0003\times10^{3}

U=199.9\times10^{-6}\ J

Hence, The electric potential energy between the two charged spheres is 199.9\times10^{-6}\ J

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When 150 joules of work is done on a system by an external force of 15 newtons in 20. seconds, the total energy of that system i
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Answer:

1.5 × 10^2 J

Explanation:

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A racetrack has the shape of an inverted cone, as the drawing shows. On this surface the cars race in circles that are parallel
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Answer:

The value of d is 183.51 m.

Explanation:

Given that,

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Suppose The car race in the circle parallel to the ground surface is at an angle 40°

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Normal force acting on the car = N

We need to calculate the value of d

Using component of normal force

The horizontal component of normal force is equal to the gravitational force.

N\cos\theta=mg....(I)

The vertical component of normal force is equal to the centripetal force

N\sin\theta=\dfrac{mv^2}{r}.....(II)

Divided equation (I) by equation (II)

\tan\theta=\dfrac{v^2}{gr}

Put the value of g

\tan\theta=\dfrac{v^2}{g\times d\cos\theta}

v^2=\tan\theta\times g\times d\cos\theta

v^2=g\times d\sin\theta

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Put the value into the formula

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3 0
3 years ago
A fence 8 ft high​ (w) runs parallel to a tall building and is 24 ft​ (d) from it. Find the length​ (L) of the shortest ladder t
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Answer:

25.3 ft

Explanation:

The illustration of the problem is shown the attached image.

The length of the ladder can be calculated using the Pythagoras theorem:

Hypotenuse^2 = opposite^2 + adjacent^2

The hypotenuse is the length of the ladder.

Hypotenuse = \sqrt{opposite^2 + adjacent^2}

L^2 = BC^2 + (24 + AE)^2........1

Triangle ABC is similar to triangle AEF, hence:

\frac{BC}{8} = \frac{AE + 24}{AE}

BC = \frac{8(AE + 24)}{AE}.................2

Substitute 2 into 1

L^2 = (\frac{8(AE + 24)}{AE})^2 + (24 + AE)^2

Let AE = x

L^2 = (\frac{8x + 192}{x})^2 + (24 + x)^2

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Minimize L with respect to x.

2\frac{dL}{dx} = 2(8 + \frac{192}{x})(-\frac{192}{x^2}) + 2(24 + x)

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5 0
3 years ago
A grinding wheel is a uniform cylinder with a radius of 8.65 cm and a mass of 0.400 kg . Calculate its moment of inertia about i
anyanavicka [17]

Answer:

I = 1.5*10⁻³ kg*m²

Explanation:

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       I = \frac{1}{2}*m*r^{2}  = \frac{1}{2}*0.400 kg*(0.0865m)^{2}  = 1.5e-3 kg*m2

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3 years ago
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