Section 8 problem 1 Point charges are shown on an axis. An observation point (fi

Question

Section 8 problem 1 Point charges are shown on an axis. An observation point (field point) is also shown. Key charge observer Charge values are shown in coulombs above each charge The axis below is in meters. 6 -3 -2 0 4 6 lulate the following: a) The electric field at the observation point. b) The force exerted on a -7 coulomb charge placed at the observation point. c) The electric potential at the observation point. d) The work required to bring a charge, -6 coulombs, to the observation point from infinity.

Explanation / Answer

Given  

a) electric field at observation point is  

q1 = -4 C is at 2 m distance and q2 = 6 C is at 5 m distance  

the field is say E1 from -4C charge and E2 from 6 C charge

we knwo that the field lines are radially outward for +ve charge

and the field lines are radially inward for -ve charge

E = kq/r^2

here field E1 is along +x axis and E2 is along -ve X axis

so net field is E = E1-E2

E1 = kq1/r1^2 , E2 = kq2/r2^2

E1 = 9*10^9*4 /2^2 = 9*10^9 N/C

E2 = kq2/r2^2  

E1 = 9*10^9*6 /5^2 = 2160000000 N/C

E = 9*10^9 - 2160000000 N/C = 6840000000 N/c along +ve x axis

b)

the force on -7 C charge at observer point due to charges q1,q2

F = kq1*q2/r^2 by coulomb's law

F = F1+F2

F = kq(q1/r1^2 +q2/r2^2)

F = (9*1^9*7)(4/2^2 + 6/5^2) N

F = 78.12 N

c) the potential at the observation point is

V = kq/r

v = v1+v2

v = k(q1/r1+q2/r2)

v = 9*10^9(-4/2 +6/5) V = -7200000000 V

d) Work required is W = ?

V = W/q

W = V*q

W = (-7200000000)(-6) J

W = 43200000000 J

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