three point charges are placed at the following points on the x-axis: +2µC at x=0, -3µC at x=40 cm, -5µC at x=120 cm. Find the force (a) on the -3µC charge (b)on the -5µC charge.
this is worst
i had type another ques and the given solu... is another 😤😤
So this is correct? Or what's the matter?
(a) Well, you know what they say, opposite charges attract! Since the -3µC charge is negative, it will be attracted towards the +2µC charge. To find the force, we can use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. So, let's calculate it:
F = (k * |q1| * |q2|) / r^2
Where:
F is the force
k is the Coulomb constant (k = 8.99 x 10^9 Nm^2/C^2)
|q1| and |q2| are the magnitudes of the charges
r is the distance between the charges
Plugging in the values:
F = (8.99 x 10^9 Nm^2/C^2) * (3 x 10^-6 C) * (2 x 10^-6 C) / (0.4)^2
Now, let's see if we can figure that out. It seems like a math workout, doesn't it?
(b) Now, let's move on to the force on the -5µC charge. Since it is negative, it will be attracted to both the +2µC and -3µC charges. So, we need to calculate the forces between these charges and sum them up.
But hey, don't worry! Math is just like a circus act - sometimes it seems complex, but in the end, it's all about having fun and using the right formulas. So, let's keep juggling those numbers and get the answer!
Remember to plug in the values in Coulomb's Law equation and calculate the individual forces between the -5µC charge and the other two charges. Afterward, simply sum them up and you'll have the total force on the -5µC charge.
I hope you're keeping track of all the numbers, because I'm juggling quite a lot here!
To find the force on a charge, we can use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, Coulomb's Law is given by:
F = k * (|q1| * |q2|) / r^2
Where:
F is the force between the charges,
k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2),
|q1| and |q2| are the magnitudes of the charges, and
r is the distance between the charges.
To find the net force on a charge due to multiple charges, we can find the individual forces on the charge due to each of the other charges and then sum them up vectorially.
(a) Force on the -3µC charge:
First, we need to find the forces due to the +2µC and -5µC charges on the -3µC charge. The distances between the charges are:
-3µC charge to +2µC charge: r1 = 40 cm
-3µC charge to -5µC charge: r2 = 80 cm
Using Coulomb's Law for the +2µC charge:
F1 = (k * |q1| * |q2|) / r1^2
= (9 x 10^9 Nm^2/C^2) * (3 x 10^-6 C) * (2 x 10^-6 C) / (0.4 m)^2
Using Coulomb's Law for the -5µC charge:
F2 = (k * |q1| * |q2|) / r2^2
= (9 x 10^9 Nm^2/C^2) * (3 x 10^-6 C) * (5 x 10^-6 C) / (0.8 m)^2
To get the net force, add the individual forces:
Net force on the -3µC charge = F1 + F2
(b) Force on the -5µC charge:
To find the force on the -5µC charge, we need to find the force due to both the +2µC and -3µC charges. The distances between the charges are:
-5µC charge to +2µC charge: r3 = 120 cm
-5µC charge to -3µC charge: r4 = 80 cm
Using Coulomb's Law for the +2µC charge:
F3 = (k * |q1| * |q2|) / r3^2
= (9 x 10^9 Nm^2/C^2) * (5 x 10^-6 C) * (2 x 10^-6 C) / (1.2 m)^2
Using Coulomb's Law for the -3µC charge:
F4 = (k * |q1| * |q2|) / r4^2
= (9 x 10^9 Nm^2/C^2) * (5 x 10^-6 C) * (3 x 10^-6 C) / (0.8 m)^2
To get the net force, add the individual forces:
Net force on the -5µC charge = F3 + F4
By calculating these equations using the given values, you should get the required values of forces on the -3µC and -5µC charges.
Fe= (9.0 * 10^9) * (-5.0 * 10^-6)(3.0 * 10^-6) / 0.80^2
= 135 * 10^-3 / 6.4 * 10^-1
= 0.2109 N
Fe= (9.0 * 10^9) * (2.0 * 10^-6)(5.0 * 10^-6) / 1.20^2
= 90 * 10^-3 / 1.44 * 10^1
= 0.0625 N
B) 0.2109375 - 0.0625 = 0.1484375
Use Coulomb's law and perform a vector addition of forces.
Fe= (-9.0 * 10^9) * (2.0 * 10^-6)(3.0 * 10^-6) / 0.40^2
= -54 * 10^-3 / 1.6 * 10^-1
= -0.3375 N
Fe= (-9.0 * 10^9) * (-5.0 * 10^-6)(3.0 * 10^-6) / 0.80^2
= 135 * 10^-3 / 6.4 * 10^-1
= -0.2109 N
a = -0.55N
Fe= (9.0 * 10^9) * (2.0 * 10^-6)(5.0 * 10^-6) / 1.20^2
= 90 * 10^-3 / 1.44 * 10^1
= 0.0625 N
Fe= (9.0 * 10^9) * (-5.0 * 10^-6)(3.0 * 10^-6) / 0.80^2
= 135 * 10^-3 / 6.4 * 10^-1
= 0.2109 N
=0.27N