## Two Small Spheres, Each Having Mass Of 0.1 G, What Is The Charge On Each Sphere. (Fsc 2Nd Year)

alarm_on09-May-2022

Today, we will solve this simple numerical in this numerical is available in the FSC second-year book. This is example. Number eleven point to the miracle is to small spheres.

Each having a mass of 0.1 gram are suspended from a point with the help of a trade 20 centimeters long. We have two small spheres. This is pure one. And this is paired to both of these piers having the same mass 0.1 gram. So the mass of each sphere is given as 0.1 gram. And these Spears are suspended from point, a with the help of.

This trip, the length of this state is given s, L is equal to 20 centimeters and 20 centimeters. We know is equal to point two zero meter. Similarly, the mass of these spheres are 0.1 gram, and we know that 0.1 gram is equal to 1 multiplied. 10 power, minus 4 kilogram dividing. This 0.1 by 1000. We will get 1 multiplied. 10 power, minus 4 K lucre.

Now, both of these Spears are equally charged, and both of them have the same charge because these charges repel each other. And the distance of separation between. These two charges is given as 24 centimeters converting this 24 centimeter in 2 meters. We will get R is equal to 0.2 4 meter dividing. This 24 centimeter by 100. We will get zero point 2, 4 meter is an SI unit. We will determine what is the charge on each pair.

They will determine the charge on spare 1 and the charge on spare 2. Now we know that the charge on spur 1 is equal to the charge on spare - q1 is equal to q2. We will devote this Q 1 and Q with Q. And we will determine the value of this Q. Now for solving this numerical, the first thing we will use is we will use the Coulomb's law according to the Coulomb's law, we know that the force of attraction or repulsion between two charges is given by F is equal to K q1 q2 by R square. This K is equal to 1 will 4 pie epsilon-not.

And the value of this term K is 9 multiplied. 10 power 9. Now, the distance of separation between these two charges is given by 0.2 4 meter. So the radius R, the distance of separation between these two charges R is equal. To zero point 2, 4 meter substituting, the values for K and R in this equation.

And we know q1 is equal to q2. So you will have HQ + HQ equal to Q Square substituting. These values, we will get the force F is equal to 9 multiplied. 10 power, 9 you square, divided by 0.2 whole square. Now, for the solving we will get finally force f is equal to 150. 6.25 multiplied. 10 power, 9 q, square.

This is equation. Number 1. We will later need this equation to determine the value of Q. Now, if we draw a line from this point. A to point B. This is point D with a midpoint of length zero point 2, 4, then we will have this length point. Point 1 2 meters, this will be 0.12 meter because this length is half of this total length.

Now we know this side is hypotenuse. This is the base. And this is the perpendicular using Pythagorean theorem. We will determine the value of this length ad.

According to primordial theorem, hypotenuse whole square is equal to B square, plus perpendicular square. The hypotenuse here is zero point. Two zero.

Meter while the base is zero point, one per meter substituting the values for hypotenuse and base. And the perpendicular is ad and solving for this ad. We will get the value of a D is equal to 0.16. Now we have determined the force F is equal to 150. Six point. Two, five.

Multiply 10 power, 9 Q, Square. This length of this side ad is 0.16 meter. Now, despair are discharged. There are four forces acting on this. Charge Q are despair force. One is the force of repulsion due to this second charge.

And the. Direction of this force is along the positive x-axis while this force F is balanced by the X component of tension T in this. Third, this tension, T have two components. One is the X component along this direction. And one is the Y component, which is along the y-axis.

The weight of this pair is acting downward. The weight of this pair is equal to mg. And the direction of this mg weight is downward. And this weight of this pair is balanced by the Y component of tension T in the third.

Similarly, the force F. Is balanced by the X component of tension T. So we have the by. The Y component of tension, T is equal to P sine theta. Similarly, the X component of tension T is given as DX is equal to P cos theta. Now, this T Y, we know is balanced by the weight of this pair, which is equal to mg.

So we have mg is equal to T sine theta. Similarly, this X component of tension, T is balanced by the force of repulsion, which is F. So we have F is equal to T cos theta now dividing this equation. Number one by equation.

Number two. We will have mg divided by F is equal to T sine, theta, divided by T cos of theta. This PE tension will cancel with this T, and we will have mg divided by F is equal to sine theta divided by cos theta. And we know that sine theta divided by cos theta is equal to tangent theta. So we have mg divided by F is equal to tangent theta. And we also know that tangent theta is equal to the Y component, divided by X. The Y component is actually the perpendicular.

And the x is the best. So we know that tangent theta. Is equal to or appendicular divided by in pass, our Y divided by X. Now here from this triangle, we can see that the perpendicular are the Y is 0.16. Similarly, the base is given as 0.1 2, substituting the values for Y and X dividing.

This 0.16 by 0.12. We will get 1 point 3 3. So we have mg divided by F is equal to 1.3 3 now solving this equation for force F substituting the mass of the spare which is given s, 1 multiplied. 10 power, minus 4 kilograms G is gravity, which is 9 point, 8 meter per second Square.

And dividing by this 1.33, we will get the value of force. F is equal to 7 point. 3 6 multiplied, 10 power, minus 4 Newton. Now we have the value of force. F is equal to 7 point, 3 6 multiplied by 10 power, minus 4 Newton. And this was the first equation which we have derived, which was F is equal to 156 point 2 5 multiplied, 10 power, 9 Q, Square, substituting, the value of this force F in this equation. Further the same defying.

This equation we will get Q square is equal to F divided by 150 6.25 multiplied. By 10 power 9 now substitute the value of F value is seven point. Three, six multiplied, 10 power, minus four in this equation.

We will have Q square is equal to seven point. Three, six multiplied by 10 power, minus four divided by 150 six point. Two, five multiplied.

10 power, 9 further, simplifying and solving we will get Q square is equal to four point. Seven multiplied, 10 power minus fifteens. Now, taking under root on both the sides, the root will cancel out with this Square, and we will have the value of Q. Charge is equal to six point. Seven multiplied, 10 power, minus eight coulombs. And this is the charge on each spear.

The charge on spear. One is six point. Seven multiplied.

10 power, minus 8 Coulombs. Similarly, the charge on the spear two is also equal to six point. Seven multiplied, 10 power, minus eight coulombs. Thank you watching and don't, forget to subscribe easy EDU for more videos. Thank you.