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Unformatted text preview: (W)y=145xlem Answers —- Week 4 In an E - ﬁeld a stationary charge q experiences a Force £5 = q E . Therefore,
attach q to the Force measuring device which can measure both magnitude and ' direction of Force. If in some region q experiences a force you know that it is
locath in an E — ﬁeld. ‘ Ex ,5.‘ Eth r2 E0) = Q + EM y—components cancel V > W
x —components add ‘ a
—2k Lines 7 '
E: eq 2 _ keqd )3
a. d2 2 %
y2 +— 2 +5]—
4 y 4
Dipole moment 13 = 610’)?
kep p j ' minim
Seem-=— ;=— 1 :3. a; 0 ft
y 47r so y 5 -~ WW? (0 q=9x10”m/s252
(ii) K =107m/sfc+l.35x105m/sfz
(iii) 1:: 0.15m)? +1v_01x1o-3m.:;~:;>»5’>’ Put Q at center of a sphere at r 91 0.
E = Q 2 p
FLUX THROUGH SPHERICAL SURFACE (DE = = "Q—
C 50 ., Mﬁwwmmr'g’ All the E ﬁeld lines must head to inﬁnity so same total ﬂux must go through any
closed surface surrounding Q. Total ﬂux is zero. 7
We DONOT KNOW E - ﬁeld at any point on surface. 4-11 ON THE SURFACE OF CONDUCTOR, UNDER STATIONARY CONDIITONS, — CHARGES MUST BE AT REST SO 5 - ﬁeld at any point inside conductor must be
zero. -13 0' = 9 x 10'10 Chm2 .1; l .1; | —175 In a conservative force work done is Independent of the path. It is determined only
by the end points. WeightEG = —M g j» is a conservative force. ' ...
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This note was uploaded on 12/28/2011 for the course PHYSICS 122 taught by Professor Bhagat during the Fall '11 term at Maryland.
- Fall '11