All categories

2 years ago

A 3.5-cm radius hemisphere contains a total charge of 6.6 × 10–7

Physics

1 answer:

spayn [35]2 years ago

4 0

As per Gauss Law

Net flux through enclosed surface is

$\phi = \frac{Q}{\epsilon_0}$

here through this hemisphere total flux will pass through two portions

1). from the curved surface

2). from flat circular base

so now we have

$\phi_{base} + \phi_{surface} = \frac{Q}{\epsilon_0}$

given that

$Q = 6.6 * 10^{-7} C$

$\phi_{surface} = 9.8 * 10^4$

now we have

$\phi_{base} + 9.8*10^4 = \frac{6.6*10^7}{8.85 * 10^{-12}}$

$\phi_{base} = - 9.8 *10^4 + 7.46 * 10^4$

$\phi_{base} = - 2.34 * 10^4 N*m^2/C$

You might be interested in

A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0

2 years ago

Determine the maximum weight of the bucket that the wire system can support so that no single wire develops a tension exceeding

stellarik [79]

Let there be N number of wires.

Maximum tension a wire can withstand = 100 lb

so, Total tension N wires can withstand = 100 N

now, total tension in N wires = Maximum weight of bucket

100 N = W

so, W = 100N

W is the weight of bucket and 100N is its maximum value.

8 0

2 years ago

What is the momentum of a 533 kg blimp moving east at +75 m/s

mylen [45]

Answer:

39975kgm/s due east

Explanation:

Given parameters:

Mass of the blimp = 533kg

Velocity = +75m/s due east

Unknown:

Momentum of the body = ?

Solution:

The momentum of a body is the amount of motion it posses.

Momentum is the product of mass and velocity;

Momentum = mass x velocity

Insert the parameters and solve;

Momentum = 533 x 75 = 39975kgm/s

The momentum of the body is 39975kgm/s due east

7 0

1 year ago

Match each force abbreviation to the correct description. Fg Fp Ff Fn force exerted by a push or pull. Support force at a right

qwelly [4]

Explanation:

Force Description

1. $F_g$ It is also known as the weight of an object. It is the force that is exerted on an object due to its mass

2. $F_p$ It is force which is exerted by a push or a pull on an object. It is also known as applied force.

3. $F_f$ It is known as resistive force. It opposes the motion of an object.

4. $F_n$ It is the force which is at a right angle to the surface or perpendicular to the surface.

3 0

2 years ago

Read 2 more answers

When θ= 0 ̊, the assembly is held at rest, and the torsional spring is untwisted. if the assembly is released and falls downward

Georgia [21]

Rod is 450mm and disk has a radius of 75mm So there is a pin holding the assembly upwards which is when Θ=0 and at the pin there is a torsional spring with constant of k=20N m/rad. One end of the rod is attached to the pin and the other is attached to the disk.

7 0

2 years ago