Physics, 02.02.2020 20:57, temperent6049

1) the electric field that is .25 m from a small sphere is 350 n/c towards the sphere. what is the charge on the sphere?

2) how much force would a 2.1x10-6 c charge feel if it was .23 m to the right of a 3.6x10-6 c charge?

3) how much power is generated in a circuit if a voltage of 9 v causes a current of .37 a?

4) a negative charge of -6.0 x 10-6 c exerts an attractive force of 15 n on a second charge that is .02 m away. what is the magnitude of the second charge?

5) sphere a is located at the origin and has a charge of 2.0x10-6 c. sphere b is located at .6 m on the x-axis and has a charge of -3.6x10-6 c. sphere c is located at .8 m on the x-axis and has a charge of 4.0x10-6 c.

a) find the net force on sphere a.

b) find the net force on sphere b.

c) find the net force on sphere c.

Answers: 1

Show answers

Answers

answered: daevinepetersopa86on

1) $2.43\cdot 10^{-9} C$

The electric field produced by a small charged sphere outside sphere is identical to that of a single point charge:

$E=k\frac{Q}{r^2}$

where

k is the Coulomb constant

Q is the charge on the sphere

r is the distance from the sphere

Here we have

E = 350 N/C is the electric field

r = 0.25 m is the distance from the sphere

Solving for Q, we find the charge:

$Q=\frac{Er^2}{k}=\frac{(350 N/C)(0.25 m)^2}{9\cdot 10^9 Nm^2 C^{-2}}=2.43\cdot 10^{-9} C$

2) 1.29 N

First of all, we need to find the magnitude of the electric field produced by the charge:

$E=k\frac{Q}{r^2}$

where here we have

$Q=3.6\cdot 10^{-6}C$ is the charge source of the field

r = 0.23 m is the distance at which we have to calculate the field

Substituting,

$E=(9\cdot 10^9 Nm^2 C^{-2}) \frac{3.6\cdot 10^{-6}C}{(0.23 m)^2}=6.12\cdot 10^5 N/C$

Now we can calculate the electric force exerted on the second charge, which is given by

F=qE

where

$q=2.1\cdot 10^{-6} C$ is the magnitude of the second charge

Substituting,

$F=(2.1\cdot 10^{-6} C)(6.12\cdot 10^5 N/C)=1.29 N$

3) 3.33 W

The power generated in a circuit is given by:

P = V I

where

V is the voltage

I is the current

In this circuit, we have

V = 9 V

I = 0.37 A

So the power generated is

P=(9 V)(0.37 A)=3.33 W

4) $+1.1\cdot 10^{-7}C$

The magnitude of the force between two electric charge is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where q1 and q2 are the two charges, and r is their separation.

In this problem we have

$q_1 = 6.0\cdot 10^{-6}C$

F=15 N

r=0.02 m

Solving for q2, we find the magnitude of the second charge

$q_2 = \frac{Fr^2}{k q_1}=\frac{(15 N)(0.02 m)^2}{(9\cdot 10^9 Nm^2 C^{-2})(6.0\cdot 10^{-6}C)}=1.1\cdot 10^{-7}C$

Moreover, the force is attractive: this means that the two charges must have opposite signs (in fact, like charges repel each other, while unline charges attract each other). In this problem, q1 is negative, so q2 must be positive.

5a) -0.07 N (towards negative x-direction)

First of all, we need to calculate the force exerted on sphere A by sphere B; this is given by

$F_{AB} = k \frac{q_A q_B}{r_{AB}^2}$

where

$q_A = 2.0\cdot 10^{-6}C\\q_B = -3.6\cdot 10^{-6}C\\r_{AB}=0.6 m$

Substituting,

$F_{AB} = k \frac{(2.0\cdot 10^{-6} C)(-3.6\cdot 10^{-6} C)}{(0.6 m)^2}=-0.18 N$

(the negative sign means the force is towards the left)

The force exerted on sphere A by sphere C is

$F_{AC} = k \frac{q_A q_C}{r_{AB}^2}$

where

$q_A = 2.0\cdot 10^{-6}C\\q_C = 4.0\cdot 10^{-6}C\\r_{AC}=0.8 m$

Substituting,

$F_{AC} = k \frac{(2.0\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.8 m)^2}=+0.11 N$

So the net force on sphere A is

$F=F_{AB}+F_{AC}=-0.18 N+0.11 N=-0.07 N$

5b) +3.06 N (towards positive x-direction)

The force exerted on sphere B by sphere A has been calculate at step 5a), and it is

$F_{AB} = k \frac{(2.0\cdot 10^{-6} C)(-3.6\cdot 10^{-6} C)}{(0.6 m)^2}=-0.18 N$

and the force is attractive, so towards the left

The force exerted on sphere B by sphere C is

$F_{BC} = k \frac{q_B q_C}{r_{BC}^2}$

where

$q_B = -3.6\cdot 10^{-6}C\\q_C = 4.0\cdot 10^{-6}C\\r_{BC}=0.8 m-0.6 m=0.2 m$

Substituting,

$F_{BC} = k \frac{(-3.6\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.2 m)^2}=-3.24 N$

and the force is still attractive, so towards the right

So the net force on sphere B is

$F=F_{AB}+F_{BC}=-0.18 N+3.24 N=3.06 N$

5c) -3.13 N (towards negative x-direction)

The force exerted on sphere C by sphere A has been calculate at step 5a), and it is

$F_{AC} = k \frac{(2.0\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.8 m)^2}=+0.11 N$

and the force is repulsive, so towards the right

Similarly, we found the force exerted on sphere C by sphere B at step 5b)

$F_{BC} = k \frac{(-3.6\cdot 10^{-6} C)(4.0\cdot 10^{-6} C)}{(0.2 m)^2}=-3.24 N$

and the force is attractive, so towards the left

So the net force on sphere C is

$F=F_{AC}+F_{BC}=+0.11+(-3.24 N)=-3.13 N$

answered: Guest

weather forecasts are done by collecting data about the current state of the atmosphere mainly the temperature, humidity and wind and using meteorology, to determine the atmosphere that evolves over time. the h and l mean the high and low pressure. the red strip is a warm front. the blue strip is a cold front. the blue and red strip is a stationary front. the purple strip is a occluded front.

explanation:

answered: Guest

answer: (431,2)

step-by-step explanation: