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devlian
1 month ago
8

Two identical conducting spheres, A and B, sit atop insulating stands. When they are touched, 1.51 × 1013 electrons flow from sp

here B to sphere A. If the total net charge on the spheres is +3.68 μC, what was the initial charge on sphere B?
Physics
1 answer:
Sav [3.1K]1 month ago
6 0
A = -0.576 μC B = 4.256 μC Suppose we consider a single electron charge. The total charge transferred from B to A is: Let A and B represent the initial charges of spheres A and B. Given that the overall charge amounts to 3.68μC, we can establish the following equation. After they come into contact, 2.416μC flows from B to A until their charges are equal, leading us to another equation. By combining both equations, we arrive at the solution.
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Two charges q1 = 5 µC, q2 = -26 µC, are L = 19 cm apart. A third charge is to be placed on the line between the two charges. How
Keith_Richards [3271]
The electric force between two objects is expressed as being proportional to the product of their charges and inversely proportional to the square of the distance separating them. In this instance, the distance between the first two charges is 19 cm. We formulate the equation k q1 q3/ (x)^2 = k q2 q3/ (19-x)^2, where x denotes the separation between q1 and q3. The charge q3 cancels out, and q2 is used in absolute terms. The resulting value of x is 5.79 cm.
6 0
1 month ago
Consider the uniform electric field \vec{E} =(4000~\hat{j}+3000~\hat{k})~\text{N/C} ​E ​⃗ ​​ =(4000 ​j ​^ ​​ +3000 ​k ​^ ​​ ) N/
kicyunya [3294]

Answer:

Electric flux is calculated as \phi=31562.63\ Nm^2/C

Explanation:

We start with the given parameters:

The electric field impacting the circular surface is E=(4000j+3000k)\ N/C

Our objective is to ascertain the electric flux passing through a circular region with a radius of 1.83 m situated in the xy-plane. The area vector is oriented in the z direction. The formula for electric flux is expressed as:

\phi=E{\cdot}A

\phi=(4000j+3000k){\cdot}Ak

Applying properties of the dot product, we calculate the electric flux as:

\phi=3000\times Ak

\phi=3000\times \pi (1.83)^2

\phi=31562.63\ Nm^2/C

Consequently, the electric flux for the circular area is \phi=31562.63\ Nm^2/C. Thus, this represents the required answer.

4 0
2 months ago
Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele
ValentinkaMS [3465]

Answer:

0.018 J

Explanation:

The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

W = q \Delta V

where

q=3.0 \mu C = 3.0 \cdot 10^{-6} C represents the charge's magnitude

and \Delta V = 6.0 kV = 6000 V signifies the potential difference between point P and infinity.

After substituting into the formula, we arrive at

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

4 0
2 months ago
d. The force is doubled and the object’s mass is halved? 18. ||| A man pulling an empty wagon causes it to accelerate at 1.4 m/s
Yuliya22 [3333]
Assuming that the mass of the empty wagon is "M," according to Newton's second law, we can derive the following relationships. Given that the empty wagon accelerates at 1.4 m/s², we proceed with this information. If a child weighing three times the mass of the wagon is on it, we can establish the relevant equations.
8 0
1 month ago
a force of 6lbs acts on an object with a weight of 35 lbs on earth. determine the objects acceleration. final answer must be 5.5
kicyunya [3294]
Weight of the object = 35 lbs
F = ma
m = F/a = 35/32 (with acceleration of 32 ft/s²)
m= 1.09

Again applying the same formula,
a = F/m
a= 6/1.09
a= 5.489

Thus, the acceleration is approximately 5.5 ft/s²!!
5 0
2 months ago
Read 2 more answers
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