Download TMA Please answer the following questions 1- 1

TMA
Please answer the following questions
1-
1-Area =2πr²+2πrh=2π(2.5) ²+2π(2.5)(6)=133.5cm²
Charge=Q/A=15x10-⁶/133.5=1.1236x10-⁷C/cm²
2-Lateral area=2πrh=2π(2.5)(6)=94.247cm²
Charge=Q/Lateral Area=15x10-⁶/94.247=1.5910-⁷C/cm²
3-Volume= πr²h= π(2.5) ² (6)=117.8cm³
Charge=Q/V=500x10-⁶/117.8=4.244C/cm³
2-
A. Three charges are at the corners of an equilateral triangle as shown in figure below .(a)
calculate the electric field at position of the 2.00µC charge due to the 7.00µC and 3.00µC.(b) use your answer to part (a) to determine the force on the 2.00µC.
E1=Ke(q/r²)=(8.99x10⁹)(7x10-⁶)/(0.5)²=251720N/C
E2=Ke(q/r²)=(8.99x10⁹)(3x10-⁶)/(0.5)²=107880N/C
Ex=E2-E1cos60=107880-251720cos60= -17980i
Ey= -E1sin60= -251720sin60= -217995.9146j
E= Ex+Ey = -17980i -217995.9146j
F=qE=(2x10-⁶)(-17980i -217995.9146j)=
-0.03596i -0.43599j N
3-
r=(1²+(0.5) ²)¹/2=1.118m
θ=cos-¹(1/1.118)=26⁰
F1=F2=ke(q1q3/r²)=(8.99x10⁹)(2x10-⁶) (3x10-⁶)/(1.118)
²
=345237 N
FX=345237cos26-345237cos26=o
FY= -2(345237sin26)= -302683.8792j
F= -302683.8792j
4-
5-
q1/q2=1:3
q1 is negative charge
q2 is positive charge
6-
A. Which of the equations below are dimensionally correct
1. v = v0 + ax
2. y=(2m) cos (kx), where k=2m-1
1-V=L/T
V0+ax=L/T+(L/T²)L
v = v + ax is not dimensionally correct
0
2-y=L
(2m) cos(kx)=kg ---}2,cos(kx) are dimensionless
y=(2m) cos (kx), where k=2m-1 is not dimensionally correct
7-
G=Fr²/Mm
=Nm²/kg²
=(kgm³/s)/kg²
=(m³/s)/kg
G= Nm²/kg² or (m³/s)/kg
Define
8-
A.
D
efine the dipole moment?
Dipole moment is the product (2aq) of the magnitude of either charge (q) and
the distance (2a) between the charges.
Dipole moment (P)=2qa
B.
D
efine the electricity?
The collection or flow of the electrons in the form of electric charge.
C.
D
efine the electric flux?
Electric field lines which penetrates a surface area A which is perpendicular to
the field.
9-
Explain how to determine the electric field of a dipole consisting of a positive
charge (q) and a negative charge (-q) separated by a distance of (2a) along the yaxis at a point (p) which is at a distance (a) from the origin.
At p the fields E1 and E2 due are equal due to the two
charges are equal in magnitude and because P is
equidistant from the two charges.
The total field is E= E1+ E2 where
E1= E2=Ke(q/r²)=ke(q/y²+a²)
The y components of E1 and E2 cancel each other
because they are in the opposite direction and the x
components of E1 and E2 add each other because they are
in the same direction.
E=2E1cosθ=2ke(q/y²+a²)(a/y²+a²)¹/2
=ke(2qa/(y²+a²)³/2)
Because y << a, we can neglect a² and write
E≈ke(2qa/y³)