1 2. Linear configured charges (6 pts): Consider the following situation where the charges are fixed. x=-2a x=-a x=0 x=+a a. What is the electric field at x=-a? b. What is the electric field at x=+2a?...

Solution
(a) Electric Field at x = -a is due to the charges at -2a , 0 and +a positions so,
Step – 1 ; Field at x=-a due to charge at x = -2a
E1 = kq/a2 ( directed towards the +ve x-axis )
Step – 2 ; Field at x=-a due to charge at x = 0
E2 = kq/a2 ( directed along –ve x-axis )
Step – 3 ; Field at x=-a due to charge at x = +a
E3 = kq/4a2 ( directed along +ve x-axis )
Hence
Net Field along + x axis
Ex = E1 + E3 = kq/4a2 + kq/a2
= 5kq/4a2
Net field along –ve x axis
E -x = kq/a2
So Total Field = Ex - E –x
= 5kq/4a2 - kq/a2
= kq/4a2
So, Electric field at x = -a is kq/4a2 (Answer)
(b) Electric Field at x = +2a is due to the charges at -2a , 0 and +a positions so,
Step – 1 ; Field at x= +2a due to charge at x = -2a
E1 = kq/16a2 ( directed towards the +ve x-axis )
Step – 2 ; Field at x=+2a due to charge at x = 0
E2 = kq/4a2 ( directed along +ve x-axis )
Step – 3 ; Field at x=+2a due to charge at x = +a
E3 = kq/a2 ( directed along - ve x-axis )
Hence ,
Net Field along + x axis
Ex = E1 + E2 = kq/16a2 + kq/4a2
= 5kq/16a2
Net field along –ve x axis
E -x = kq/a2
So Total Field = Ex - E –x
= 5kq/16a2 - kq/a2
= -11kq/16a2
So, Electric field at x = +2a is 11kq/4a2 directed along –ve x – axis (Answer)
(c) For calculating Force on charge of magnitude 2q placed at x = -a we can use the field present at that point due to
all other Charges.
We know that a charge experiences force whenever placed in Electric Field given by;
F = qE
= 2q x Electric field at x=-a
= 2q x kq/4a2
= kq2/2a2 (Answer)
Solution
(a) For calculating force on 1nC of charge we need to Calculate Electric Field at 1nC of Charge.
Step – 1 Calculation of Dipole Moment of the Dipole
Since 3nC and -3nC of charge are forming a dipole then their Dipole Moment can be calculated as follows,
P = 2l x q = 30 x 10-2 x 3 x 10-9 = 0.9 x 10-9 Cm
Step – 2 Calculation of Electric Field due to dipole on its Equitorial Line
Since 1 nC is present on the Equitorial line(
oad – on positions ) of the dipole hence, distance of 1nC of charge from the
centre of the dipole
= √3
2
?? ( length of the altitude of the equilateral triangle)
= 25.98cm = 26cm
Now Electric field on its equitorial line is
E = k p/(r2 + l2)3/2 where l is half length of the dipole
Since p , r and l all the three quantities are known so we have
On replacing the values of quantities it gives
E = 3.00 x 102 N/C
Step – 3 Calculation of Force on 1nC of Charge
F = qE
= 1 x 10-9 x 3.0 x 10 2
F = 3.0 x 10-7 N along the base towards the -3nC of charge (Answer)
(b) Electric Field at the center of the triangle
Since center of the triangle in an equilateral triangle also acts as Circumcenter of the triangle hence distance of center
from all the three Charges will be
= √3
3
a = 17.32cm
Now , Electric Field due to 1nC of Charge
E1 = kq
2
On replacing values of quantities we get
= 300 N/C , directed away from the charge along the altitude
Electric Field due to 3nC of Charge
E2 = kq
2
On replacing values of quantities we get
= 900 N/C , directed away from the charge along the altitude
Electric Field due to -3nC of Charge
E3 = kq
2
On replacing values of quantities we get
= 900 N/C, directed towards the charge along the altitude
Now the resultant field due to E1 and E2 is
E12 = sqrt(E12 + E12 - 2 E1 E2Cos1200 )
= 1081.66 N/C
Since E12 is directed in the direction of E3 so the net electric field at the center of triangle is
E = E12 + E 3
= 1981.66 N/C directed along -3.0 nC of charge.(Answer)
(c) For calculating Electric Field at the mid point of the two charges we need to calculate fields due to all the three
charges
Step-1 Field due to +3nC of Charge
E1 = kq
2
= 9 x 109 x 3 x 10-9 / .152
= 1200 N/C along the side towards 1nC of Charge
Step-2 Field due to + 1nC of Charge
E2 = kq
2
= 9 x 109 x 1 x 10-9 / .152
= 400 N/C along the side towards 3nC of Charge
Step-3 Field due to -3 nC of Charge
E3 = kq
2
= 9 x 109 x 1 x 10-9 / .262
= 133 N/C along the altitude towards -3nC of Charge
Now net electric field due to E1 and E2 is 800 N/C since they are directed opposite to each other and directed along the
+3nc of charge.
E12 = 800 N/C
Since E12 and E3 are perpendicular to each other so net electric field at mid point is
E = sqrt(E122 + E32)
On calculating
E = 810.89 N/C at the mid point directed at an sngle 450 w.r.t +3nc of charge inwards the triangle(answer)
Solution
For calculating electric field due to a point charge using Gauss’s law we need to consider a gaussian sphere of radius r
(the distance at which field Is to be calculated due to point charge.)
Let charge Q is present at the center of the gaussian sphere of radius r. So from Gauss’s law we have
Where A is the area of the spherical gaussian surface and E is the electric field due to the point charge,
On the other hand, Gauss' law shows
Where Q is the charge...