Homework+Ch+21

# Homework Ch 21

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ty + Fg = T cos θ − mg = 0 mg cos θ T= Now let’s look at the horizontal direction Fnet,x = Tx + Fe = −T sin θ + Fe = 0 3 θ l E T Q,m d Fe Fg √ l2 − (l − d)2 2dl − d2 = −mg l−d l−d Fe = T sin θ = mg tan θ = mg √ 2 mg 2ldl−d Fe −d = Q= E E √ 2 −3 )(9.8) (2)(0.12)(0.55)−0.12 (10 0.55−0.12 = 15000 = 5.21 × 10−7 C We know that Fe must be towards the right to balance the force, so the charge must have been positive. 21.35 E = E+Q + E−Q = kQ kQ 4kQax − =− 2 2 (x + a) (x − a) (x + a)2 (x − a)2 The negative sign indicates that the ﬁeld is to the left. 21.46 Consider a inﬁnitesimal piece dy of the wire, at the position +y, the ﬁeld it creates at point P is: 1 dQ λdy 1 dE = ˆ= r ˆ r 2 2 + y2 4π 0 r 4π 0 x 4 dy r y θ x P By symmetry, the vertical component of the ﬁeld will cancel out with the vertical component of the ﬁ...
View Full Document

## This note was uploaded on 09/11/2012 for the course PHYSICS 002 taught by Professor Welsch during the Spring '12 term at University of California, Berkeley.

Ask a homework question - tutors are online