Four point charges qA = 2 μC, qB = – 5 μC, qc = 2 μC, and qD = –5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

Suppose a square ABCD with each side of 10 cm and centre 0. At the centre, the charge of 1 μC is placed.

Suppose a square ABCD with each side of 10 cm and centre 0. At the ce
straight q subscript straight D space equals space minus 5 space μC
straight q subscript straight C space equals space 2 space μC
straight q subscript straight A space equals space 2 space μC
straight q subscript straight B space equals space minus 5 μC
As qA = qc, the charge of 1 μC experiences equal and opposite forces FA and Fc due to charges qA and qC.
At the same time, the charge 1 μC experiences equal and opposite forces. FB and FD due to equal charges qB and qD at B and D. Thus, the net force on charge of 1 μC due to the given charges is zero.
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Two point charges qA = 3 μC and qB = –3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 x 10–9 C is placed at this point, what is the force experienced by the test charge?


                          straight q subscript straight A space equals space 3 space μC space equals space 3 space cross times space 10 to the power of negative 6 end exponent straight C
straight q subscript straight B space equals space minus 3 space μC space equals space minus 3 space cross times space 10 to the power of negative 6 end exponent straight C
and                        straight d space equals space 20 space cm
(a) 
         
                          and                 ?
Let us assume that a unit positive test charge is placed at 0. qA will repel this test charge while qB will attract. Hence, stack straight E subscript 1 with rightwards arrow on top space and space stack straight E subscript 2 with rightwards arrow on top both are directed towards OB with rightwards arrow on top.
therefore                           straight E space equals space stack straight E subscript 1 with rightwards arrow on top space plus space stack straight E subscript 2 with rightwards arrow on top

                               box enclose straight E equals fraction numerator 1 over denominator 4 πε subscript 0 end fraction straight q over straight r squared end enclose
                               
                                     equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction. straight q subscript straight A over straight r squared plus fraction numerator 1 over denominator 4 πε subscript 0 end fraction straight q subscript straight B over straight r squared space equals fraction numerator 1 over denominator 4 πε subscript 0 straight r squared end fraction open square brackets straight q subscript straight A plus straight q subscript straight B close square brackets
equals space fraction numerator 9 cross times 10 to the power of 9 over denominator left parenthesis 0.1 right parenthesis squared end fraction open square brackets 3 cross times 10 to the power of negative 6 end exponent plus 3 cross times 10 to the power of negative 6 end exponent close square brackets
equals space 5.4 space cross times space 10 to the power of 6 space NC to the power of negative 1 end exponent space along space OB.
(b) As a negative test charge of q0 = – 1.5 x 10–6 C is placed at 0. qA will attract it while qB will repel. Therefore, the net force

                          and                 ?

straight F space equals space straight F subscript 1 plus straight F subscript 2
box enclose straight F equals fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight q subscript 1 straight q subscript 2 over denominator straight r squared end fraction end enclose
straight F equals open vertical bar straight F subscript 1 close vertical bar plus open vertical bar straight F subscript 2 close vertical bar
space space space equals space fraction numerator Kq subscript straight A straight q subscript 0 over denominator straight r squared end fraction plus fraction numerator Kq subscript straight B straight q subscript 0 over denominator straight r squared end fraction
space space space equals space straight K subscript straight q subscript 0 end subscript over straight r squared open square brackets straight q subscript straight A plus straight q subscript 0 close square brackets
space space space equals space fraction numerator 9 cross times 10 to the power of 9 cross times 1.5 cross times 10 to the power of negative 19 end exponent left square bracket 3 cross times 10 to the power of negative 6 end exponent plus 3 cross times 10 to the power of negative 6 end exponent right square bracket over denominator left parenthesis 0.1 right parenthesis squared end fraction
   equals space fraction numerator 9 cross times 1.50 cross times 6 cross times 10 to the power of negative 6 plus 9 minus 9 end exponent over denominator 0.1 cross times 0.1 end fraction
equals space fraction numerator 9 cross times 10 to the power of 9 cross times 3 cross times 10 to the power of negative 6 end exponent cross times 1.5 cross times 10 to the power of negative 9 end exponent over denominator left parenthesis 0.1 right parenthesis squared end fraction plus fraction numerator 9 cross times 10 to the power of 9 cross times 3 cross times 10 to the power of negative 6 end exponent cross times 1.5 cross times 10 to the power of negative 9 end exponent over denominator left parenthesis 0.1 right parenthesis squared end fraction
equals space fraction numerator 9 cross times 10 to the power of 9 cross times 3 cross times 10 to the power of negative 6 end exponent cross times 1.5 cross times 10 to the power of negative 9 end exponent cross times 2 over denominator left parenthesis 0.1 right parenthesis squared end fraction
equals space 8.1 space cross times space 10 to the power of negative 3 end exponent straight N.


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Check that the ratio ke2Gmempis dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?


                       Electrostatic force is given by F = Kq1q2r2Gravitational force is given by F =Gm1m2r2
                      where G = 6.67 × 10-11N m2 kg-2            me = 9.1 × 10-31 kg             m p = 1.67 × 10-27 kg            e = 1.6 × 10-19 C
and   k=14πε0 = 9×109 Nm2C2
Now,       ke2Gmemp = 9×109Nm2C2×1.6×10-19C×1.6×10-19C6.67×10-11Nm2kg-2×9.1×10-31kg×1.67×10-27kg                 = 2 × 27 × 1039 which is dimensionless.

It also establishes that the electrostatic force is about 10
39 times stronger than the gravitational force.

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When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Law of conservation of charge states that total charge on an isolated system of objects always remain conserved.
When a glass rod is rubbed with silk cloth, glass rod becomes positively charged while silk cloth becomes negatively charged and, the amount of positive charge on the glass rod is apparently found to be exactly the same as the negative charge on silk cloth. Since, the measure of charge is same on both, equal amount of charge with opposite nature will cancel out each other. Hence, the total sum of charge on two bodies is zero. Thus, the system of glass rod and silk cloth, which was neutral before rubbing, still possesses no net charge after rubbing. And law of conservation of charge is justified. 

As a consequence of conservation of charge, when two charged conductors of same size and same material carrying charges Q1 and Q2 respectively are brought in contact and separated, the charge on each conductor will be Q1+Q22.
This condition, however, does not hold true if the conductors are of different sizes or of different material.
In that case the charges on the conductors will be Q1and Q2’ respectively, where Q1 + Q2 = Q1‘ + Q2’.

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A system has two charges qA = 2.5 x 10–7 C and qB = –2.5 x 10–7 C located at points A: (0, 0, –15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?

Total charge,

straight q equals straight q subscript straight A plus straight q subscript straight B
space space equals space 2.5 space cross times space 10 to the power of negative 7 end exponent space minus space 2.5 space cross times space 10 to the power of negative 7 end exponent space equals space 0


Total charge,a = AB = 15 – (– 15) = 30 cm = 0.3 m Electric dipole

a = AB = 15 – (– 15) = 30 cm = 0.3 m Electric dipole moment
box enclose straight p space equals space straight q. space straight a end enclose
equals 2.5 space cross times space 10 to the power of negative 7 end exponent space left parenthesis 0.3 space straight m right parenthesis
equals space 7.5 space cross times space 10 to the power of negative 8 end exponent space Cm space left parenthesis along space space minus straight Z minus axis right parenthesis.
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