Physics, 21.07.2021 20:30 fireemblam101ovu1gt

Two small spheres assumed to be identical conductors are placed at 30 cm from each other on a horizontal axis. the first S1 is loaded with 12 nC and the second S2 is loaded with -18 nC. make a figure where you need representative. 1) determine the electric force exerted by S1 on S2
b) determine the electric field created by S1 at the level of s2
c) determine the electric potential created by s1 at the level of s2
d) determine the electric force exerted by S2 on S1
e) determine the electric field created by s1 in the middle of [S1: S2]
f) determine the electric field in the middle of [S1: S2]
g) determine the electric potential in the middle of [S1: S2]
h) determine the electric potential at the level of S1

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The electric force exerted by S1 on S2 is 21.58μN. The electric field created by S1 at the level of S2 is $1.20 \frac{kN}{C}$ . The electric potential created by S1 at the level or S2 is 360V. The electric force exerted by S1 on S2 is 21.58μN. The electric field generated by S1 in the middle of S1 and S2 is $4.79 \frac{kN}{C}$ . The electric field in the middle of S1 and S2 is $11.99 \frac{kN}{C}$ . The electric potential in the middle of S1 and S2 is 1.80 kV. The electric potential generated by S2 on the position of S1 is 539.4V

a) The electric force exerted by S1 on S2 is 21.58μN.

b) The electric field created by S1 at the level of S2 is $1.20 \frac{kN}{C}$

c) The electric potential created by S1 at the level or S2 is 360V

d) The electric force exerted by S1 on S2 is 21.58μN.

e) The electric field generated by S1 in the middle of S1 and S2 is $4.79 \frac{kN}{C}$

f) The electric field in the middle of S1 and S2 is $11.99 \frac{kN}{C}$

g) The electric potential in the middle of S1 and S2 is 1.80 kV

h) The electric potential generated by S2 on the position of S1 is 539.4V

The magnitude of the force is determined by using the following formula:

$F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}$

where:

F= Electric force [N]

k = Electric constant ( $N\frac{m^{2}}{C^{2}}$ )

q1= First charge [C]

q2 = Second charge [C]

r = distance between the two charges

So, in this case, the force can be calculated like this:

$F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}$

So the force will be equal to:

$F=21.58x10^{-6}N$

which is the same as:

$F=21.58 \micro N$

b) The electric field created by S1 at the level of S2 is $1.20 \frac{kN}{C}$

We can take the following formula for the electric field.

$E_{S1}=\frac{F_{e}}{q_{2}}$

$E_{S1}=\frac{1.20 x10^{-6}N}{18x10^{-9}C}$

which yields:

$E_{S1}=1.20x10^{3} \frac{N}{C}$

$E_{S1}=1.20 \frac{kN}{C}$

In this case, since S1 is positive, we can asume the electric field is in a direction away from S1.

The electric potential created by S1 at the level of S2 is 360V

This electric potential can be found by using the following formula:

V=Er

Where V is the electric potential and it is given in volts.

So we can use the data found in the previous sections to find the electric potential:

$V=(1.20x10^{3} \frac{N}{C})(30x10^{-2}m)$

V=360V

d) The force exerted by S2 on S1 will be the same in magnitude as the force exerted by S1 on S2 but oposite in direction. This is because the force will depend on the two charges, and the distance between them, so:

The electric force exerted by S1 on S2 is 21.58μN.

The magnitude of the force is determined by using the following formula:

$F_{e}=k_{e}\frac{|q_{1}||q_{2}|}{r^{2}}$

$F_{e}=(8.99x10^{9}N\frac{m^{2}}{C^{2}})\frac{|12x10^{-9}C||18x10^{-9}C|}{(30x10^{-2}m)^{2}}$

So the force will be equal to:

$F=21.58x10^{-6}N$

which is the same as:

$F=21.58 \micro N$

e) The electric field generated by S1 in the middle of S1 and S2 is $4.79 \frac{kN}{C}$

In order to find the electric field generated by S1, we can make use of the following formula

$E=k_{e} \frac{q_{1}}{r_{1}^{2}}$

$E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})(\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}})$

which yields:

$E=4.79 \frac{kN}{C}$

f) The electric field in the middle of S1 and S2 is $11.99 \frac{kN}{C}$

In order to find the electric field generated by two different charges at a given point is found by using the following formula:

$E=k_{e} \sum \frac{q_{i}}{r_{i}^{2}}$

where:

$q_{i}$ = each of the charges in the system

$r_{i}$ = the distance between each of the charges and the point we are analyzing.

We'll suppose the electric fields will be positive then, so:

$E=(8.99x10^{9} N\frac{m^{2}}{C^{2}})[\frac{12x10^{-9}C}{(15x10^{-2}m)^{2}}+\frac{18x10^{-9}C}{(15x10^{-2}m)^{2}}]$

which yields:

$E=11.99 \frac{kN}{C}$

g) The electric potential in the middle of S1 and S2 is 1.80 kV

Since we know what the electric field is from the previous question, we can make use of the same formula we used before to find the electric potential in the middle of S1 and S2

So let's take the formula:

V=Er

So we can use the data found in the previous sections to find the electric potential:

$V=(11.99x10^{3} \frac{N}{C})(15x10^{-2}m)$

V=1.80kV

The electric potential generated by S2 on the position of S1 is 539.4V and can be found by using the following formula:

$V=k_{e}\frac{q_{2}}{r}$

So we can use the data provided by the problem to find the electric potential.

$V=(8.99x10^{9} N\frac{m^{2}}{C^{2}})(\frac{18x10^{-9}C}{30x10^{-2}m})$

V=539.4V

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