PHYSICS HELP PLEASE!

Question

A large electroscope is made with "leaves" that are 78-cm-long wires with tiny 47 g spheres at the ends. When charged, nearly all the charge resides on the spheres.

 

If the wires each make a 26° angle with the vertical, what total charge Q must have been applied to the electroscope? Ignore the mass of the wires.

Answer 1

   ↑  \

   ↑   \

   ↑    \

   ↑     \

   ↑26° \

   ↑       \ 

   ↑        \

   ↑         \

F_y↑          \ F_Tension and length = 78cm

   ↑           \

   ↑            \ 

   ↑     x       \ Q₂  (mass = 49g)

    ←←←←← O →→→→→ F_q  = electrostatic force of repulsion

        F_x         |

                    |

                    |

                    ↓

                    F_g = weight

 

From the free body diagram, we can resolve the force due to tension, F_T ,  into the x and y components, F_x and F_y.

We can also say that F_y = F_g (weight) and F_x= F_q (electrostatic force).

The distance x is one half of the distance separating the two charged spheres. So d = 2x.

Use SOHCAHTOA to solve for the distance x and to solve for the components F_x and F_y. Then, use Coulomb's law to solve for the charge.

SOLUTION:

F_g = F_y = mg = (.047kg)(9.8m/s²) = .4606 newtons

F_y = .4606 newtons

 

tan 26° = Opp/Hyp = (F_x / F_y) ⇒ F_x = F_y tan26° = .2246 newtons

F_x = F_q = .2246 newtons

 

sin26° = Opp/Hyp = (x / 78cm) ⇒ x = 78 cm sin26° = 34.1929 cm

Since x is only half of the distance between the spheres, then multiply by 2 to get the distance d.

d = 68.3858 cm or 0.683858 m

 

F_q = k (Q₁ Q₂) / d²  and since Q₁ = Q₂ so,

 

F_q  = k (Q² / d²)    ⇒   Q² =   F_q d² / k

 

Q² = (.2246N)(0.683858m)²/ (9 x 10⁹ Nm²/C²)

 

Q² = 1.1671 x 10^-11 C²

 

Q = 3.42 x 10^-6 C or 3.4 x 10^-6 C with sig figs.

Answer 2

the spheres at 26 deg form a pendulum, so the weight of the sphere in x direction is 47 times g times tan 26

convert to mks

I get 22.46 X 10 -2 newtons

since this is half the bottom of the triangle formed by the leaves, we have to double this to get the equilibrium force

so 44.92 X 10 -2 N

the dist between two charges is 2 times sin 26 times .78 m or .64 m

 

now the equation for electric force between two charges is F =  Kc (qq) / r2

 

the force holding the spheres apart is the value above, r is equal to the twice the .78 m times sin 26 or .68 m 

 

q is square root of r2 times F divided by kc , substituting I get 4.51 X 10-6 Coulombs