A ball of mass m=1 kg and charge Q=2.50E-4 C hangs (at sea level) from a massles

Question

A ball of mass m=1 kg and charge Q=2.50E-4 C hangs (at sea level) from a massless thread of length L=1.8 m from a pole of linear charge density =5.00E-7 C/m. Disregard the radius of the ball and pole; assume that the ball hangs from a ring that allows free rotation about the pole; and treat the pole as infinitely long. What is the angle (in degrees) of the thread relative to the line of charge?

A ball of mass m=1 kg and charge Q=2.50E-4 C hangs (at sea level) from a massless thread of length L=1.8 m from a pole of linear charge density ?=5.00E-7 C/m. Disregard the radius of the ball and pole; assume that the ball hangs from a ring that allows free rotation about the pole; and treat the pole as infinitely long. What is the angle ? (in degrees) of the thread relative to the line of charge?

Explanation / Answer

We can calculate the electric field coming from the pole from gauss' law.

i.e integral E ds = qenclosed/0 , ds is any cylindrical shell concentric with the pole.

From here we get E= /20r where r is the radius from the pole,

and is the linear charge density and is given to be 5.0x10-7C/m   

Therefore now the forces acting on the ball are Tension , mg (gravity) and qE (electric force).

We get two equations, Tcos = mg --------------------- (1)

Tsin = qE -------------- (2)

From (1)& (2),

==> tan = qE/mg

---> tan = q/20rmg

Now r is the perpendicular distance from the pole ----> i.e r is Lsin

Therefore tan = 2 x 5.0x10-7C/m x 2.4 x 10-4C/(1.8*9.8 x 40)sin

--> if we substitute cos = t

then we get (1-t2)/t = 1.36x10-2

----> t2 +0.0136t -1 = 0

-----> t =0.9932

----> = 6.68 degrees.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.