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3 years ago

A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.

Physics

1 answer:

sladkih [1.3K]3 years ago

5 0

Explanation:

Given that,

Charge 1, $q_1=6.75\ nC=6.75 \times 10^{-9}\ C$

Charge 2, $q_2=4.46\ nC=4.46\times 10^{-9}\ C$

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

$F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}$

$F=6.84\times 10^{-8}\ N$

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

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3 years ago

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A point charge q1=2.0μC is located on the positive y axis at y=0.30m, and an identical charge q2 is at the origin. Find the magn

dalvyx [7]

Answer:

(A) 0.279N at angle 38.02°

(B) 0.701N

Explanation:

(A) The net force on q3 is given as:

F = Fxi + Fyj

Fx is the x component of the force

Fy is the y component of the force

Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0

Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx

First let us find y and angle x from the diagram.

Using Pythagoras theorem,

y² = 0.3² + 0.4²

y² = 0.25

y = 0.5m

Using SOHCAHTOA to find x,

sinx = 0.4/0.5

x = 53.13°

Electrostatic force, F is given as:

F = kqQ/r²

Where k = Coulumbs constant

F(1,3) = (k*q1*q3) / r²

F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)

F(1, 3) = 0.288N

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = 0.45N

Therefore,

Fx = -0.288cos36.87 + 0.45

Fx = 0.22N

Fy = 0.288cos53.13

Fy = 0.172N

=> F = 0.22i + 0.172j

The magnitude of the force will be

F(mag) = √(0.22² + 0.172²)

F(mag) = 0.279N

The direction of the force makes will be

tanθ = Fy/Fx

tanθ = 0.172/0.22 = 0.781

θ = 38.02° to the x axis.

(B) q2 = - 2.0 * 10^(-6)

This implies that:

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = -0.45N

Therefore,

Fx = -0.288cos36.87 - 0.45

Fx = -0.68N

Fy = 0.172N

=> F = - 0.68i + 0.172j

The magnitude of the force will be

F(mag) = √((-0.68)² + 0.172²)

F(mag) = 0.701N

tanθ = 0.172/0.68

θ = 14.19° to the x axis

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Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou

kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

i = current flow = 3 A
t = time interval for which the current flow = $4\ h = 4\times 3600\ s = 14400\ s$
V = terminal voltage of the battery
R = rate of energy = 9 cents/kWh

<u>Assume:</u>

Q = charge transported as a result of charging
E = energy expended
C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

$\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\$

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

$E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\$

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

$\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents$

Hence, 0.108V cents is the charging cost of the battery.

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3 years ago