Download PH504-10-test-Q-and-A

PH504/10/Test
UNIVERSITY OF KENT
SCHOOL OF PHYSICAL SCIENCES
CLASS TEST PH504-A
Friday 12th November 2010
These questions will be marked each out of 25. Answering the four questions should take 40
minutes). Question S4 is printed overleaf.
S1. In some region of space, the electrostatic potential is the following function of Cartesian
coordinates x, y, and z: V(x,y,z) = x2 + 2xy,
where the potential is measured in volts and the distances in meters. Find the electric field at the
point x = 2 m, y = 2 m, z = 2m. .
[25 marks]
S2. State how capacitance C, charge Q and potential V are related.
[10 marks]
Determine the capacitance of a single capacitor that will have the same effect as the
combination of 3 shown here .
[15 marks]
S3. If the above system with C1 = 2 F, C2 = 1 F, and C3 = 3 F are connected to a 24-V
source, shown as V, calculate the charge stored on each capacitor.
[25 marks]
S4. A horizontal wire with a mass per unit length of 0.2 Kg/m carries a current of 4 A in the
+x-direction. If the wire is placed in a uniform magnetic flux density B , determine the
direction and minimum magnitude of B in order to magnetically lift the wire vertically
upward. Please write the result in vectorial form and use a diagram to show the vectors
involved.
[25 marks]
(Hint: The acceleration due to gravity is g  9.8k m/s2, where k is the unit vector oriented
along the Z-axis.
S1. The electric field is the gradient of -V
The gradient can be expressed vector form e.g.
V  (V /x,V /y,V /dz)
V  (2x  2y,2x,0)



E  (8,4,0)
S2. C = Q/V
The voltage Vp across capacitors in parallel 2 and 3 is the same. The
charge stored is added. Therefore, the capacitance of the parallel
component is
Cp = (Q2 + Q3)/Vp
= C 2 + C3
The charge, Q, on C1 and Cp is the same. The voltage is now additive.
given by V = V1 + Vp
Therefore, the capacitance across both C1 and Cp is given by
1/C = 1/C1 + 1/Cp
= 1/C1 + 1/(C2 + C3)
S3. Substituting C = 4/3 F
So Q = CV = 32 C
Charge stored Q =Q1 = Q2+Q3
So Q1 = 32 C
V1 = Q1/C1 = 16 V
So Vp = 24V – 16V = 8 V
Q2 = C2 x Vp = 8C
Q3 = C3x Vp = 2
S4 Solution
Marks
For a length l
F  k0.2l  9.8  k1.96l ( N) (For G = mg, 2 points)
6
F m  iIlx B
8
If only lIB and no vector, 6 points ( 2points for vector notation).
In order to compensate the gravitation force, the magnetic force on the wire has to
be oriented in the vertical plane, upwards. According to to the rules of the vectorial
product, vector B is along the positive y direction, i.e. B enters the paper..
Z
Y
6
I
+
Y
1.96l = IlB
so B = 1.96/4=0.49 T
and B  0.49 j
X
2
1
2
25