Part A Use Coulomb's law to determine the magnitude of the electric field at points A and B in (Figure 1) due to the two positive charges (Q = 4.9 µC ) shown. Suppose that a = 5.9 cm. Determine the magnitude of the electric field at A Express your answer to two significant figures and include the appropriate units. HA ? EA = Value Units Submit Request Answer Figure < 1 of 1 Part B Determine the angle between the direction of the electric field at point A and the positive x-direction. В Express your answer using two significant figures. +Q• +Q μVα ΑΣφ a a 2a Submit Request Answer
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Water constitutes about 70% of earth. Some important distinguishing properties of water are high molar concentration, small dissociation constant and high dielectric constant.
Electrostatic Potential and Capacitance
An electrostatic force is a force caused by stationary electric charges /fields. The electrostatic force is caused by the transfer of electrons in conducting materials. Coulomb’s law determines the amount of force between two stationary, charged particles. The electric force is the force which acts between two stationary charges. It is also called Coulomb force.
Given,
a = 5.9 cm
Q = 4.9 micro coulomb
Magnitude of Electric field at point A due to the left charge Q will be
EA = (1/4πeo)(Q/r2 )
here r is the distance between the charge Q and the point A which can be calculated using Pythagoras theorem.
r =√{(2a)2+(a)2}
r = √5 a
EA = (9*109 * 4.9*10-6)/(5*5.9*5.9*10-4)
= 0.25 * 107 N/C
= 2500000 N/C
Since the other charge is also at the same distance and had the same magnitude so the Electric field due to it will also be equal to above value.
E'A = 2500000 N/C
These two Electric fields are symmetric to each other . Their vertical components will add up whereas the horizontal component will cancel each other as they are in opposite directions.
The angle between the Electric field and the positive x direction for the left charge is
Angle N = tan-1(1/2) = 26.565°
The sine components of the Electric fields due to both charges will add up vertically
Net Electric field at A
= 2*2500000*sin(26.565°)
= 2236063.98 N/C
In 2 significant figure, Electric field is
2.2 * 106 N/C
In 2 of figure, Angle between Electric field at A and the positive x direction is
Angle N = 90°
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