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4 months ago

A sphere of radius 0.03m has a point charge of q= 7.6 micro C located at it’s centre. Find the electric flux through it?

Physics

1 answer:

Sav [3.1K]4 months ago

7 0

Answer:

The electric flux going through the sphere is $8.58 *10^{5} \frac{Nm^2}{C}$

Explanation:

Given

$Radius,\ r = 0.03m\\Charge,\ q =7.6\µC$

Required

Calculate the electric flux

Electric flux can be computed using the formula;

Ф = q/ε

Where ε stands for the electric constant permittivity

ε $= 8.8542 * 10^{-12}$

Substituting ε $= 8.8542 * 10^{-12}$ and $q =7.6\µC$ ; the formula simplifies to

Ф = $\frac{7.6\µC}{8.8542 * 10^{-12}}$

Ф = $\frac{7.6 * 10^{-6}}{8.8542 * 10^{-12}}$

Ф = $\frac{7.6}{8.8542} *\frac{10^{-6}}{10^{-12}}$

Ф = $\frac{7.6}{8.8542} *10^{12-6}}$

Ф = $0.85834970974 *10^{12-6}}$

Ф = $0.85834970974 *10^{6}}$

Ф = $8.5834970974 *10^{5}}$

Ф = $8.58 *10^{5} \frac{Nm^2}{C}$

Thus, the electric flux through the sphere is $8.58 *10^{5} \frac{Nm^2}{C}$

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Answer:

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4 months ago

An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.51 μC/m2. Another infini

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3 months ago

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m

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Answer:

(A) The tension's magnitude grows to four times the initial value, 4F.

Explanation:

When an object travels in a circular path, a centripetal force is exerted upon it. In this instance, the centripetal force acting on the stone can be represented by $\frac { m{ v }^{ 2 } }{ r }$ .

Here, m denotes the mass of the object

v is the velocity or speed of the object

r signifies the radius of the circular path

Importantly, the tension corresponds to the centripetal force.

Initially, the string completes one revolution each second, and subsequently, it accelerates to perform two revolutions in the same time frame. This signifies that the speed has increased twofold.

Applying our formula:F = $\frac { m{ v }^{ 2 } }{ r }$

where F indicates the tension in the string

assuming the starting speed is v, after doubling it becomes 2v

Maintaining the circle's radius, we arrive at:

F= $\frac { m{ (2v) }^{ 2 } }{ r } =\frac { 4m{ v }^{ 2 } }{ r }$

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2 months ago

(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i

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Response:

a) $P = 370.993\,kPa$ , b) $F = 25.948\,kN$

Clarification:

a) The absolute pressure at a depth of 27.5 meters is:

$P = P_{atm} + P_{man}$

$P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)$

$P = 370.993\,kPa$

b) The force applied by the water is:

$F = (P - P_{atm})\cdot A$

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3 months ago

Read 2 more answers

A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of

Sav [3153]

Answer:

A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.

Explanation:

τchild=τrock

We will utilize the formula for torque:

(F)child(d)child)=(F)rock(d)rock)

The gravitational force acts equally on both objects.

(m)childg(d)child)=(m)rockg(d)rock)

We can eliminate gravity from both sides of the equation for simplification.

(m)child(d)child)=(m)rock(d)rock)

Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.

(25kg)(1m)=(50kg)drock

Solve for the distance where the rock should be positioned in relation to the seesaw's center.

drock=25kg⋅m50kg

drock=0.5m

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3 months ago