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Two point charges 2 C and 3 uC are placed at two comers of an equilateral triangle of side 20 cm in free space. Calculate the magnitude of resultant electric field at the third corner of the triangle. If an & - particle is placed at the third corner, what is the force acting on it? (Charge on & – particle is 3.2 x 10-19 €).
Asked by vedhachethan | 19 Sep, 2021, 01:20: PM
answered-by-expert Expert Answer
It is assumed the point charges are 2 μC and 3 μC .
 
Particle placed at third vertex is α particle of charge 3.2 × 10-19 C
-------------------------------------
Figure shows charge of 2 μC and 3 μC respectively at the vertex A and B of equilateral triangle .
 
Let us choose C as the origin of cartesian coordinate system as shown in figure.
 
Magnitude of electric field EA at C due to 2 μC charge at A is given as
 
begin mathsize 14px style open vertical bar E subscript A close vertical bar space equals space K space q over d squared space equals space 9 space cross times space 10 to the power of 9 space cross times fraction numerator 2 cross times 10 to the power of negative 6 end exponent over denominator 0.2 space cross times 0.2 end fraction space equals space 4.5 space cross times space 10 to the power of 5 space end exponent N divided by C end style
Where K = 1/(4πεo ) = 9 × 109 N m2 C-2 is Coulomb's constant
 
Magnitude of electric field EB at C due to 3 μC charge at A is given as
 
begin mathsize 14px style open vertical bar E subscript B close vertical bar space equals space space 9 space cross times space 10 to the power of 9 space cross times fraction numerator 3 cross times 10 to the power of negative 6 end exponent over denominator 0.2 space cross times 0.2 end fraction space equals space 6.75 space cross times space 10 to the power of 5 space end exponent N space divided by space C end style
Resultant field E due to both charge is wriiten as by consideing above fields in vector form
 
E = EA + EB  =  ( | EA |  + | EB | ) cos30 begin mathsize 14px style i with hat on top end style + ( | EB |  - | EA | ) sin30 begin mathsize 14px style j with hat on top space end style  
E = ( 9.743 begin mathsize 14px style i with hat on top end style + 1.125 begin mathsize 14px style j with hat on top space end style) × 105  N / C
Magnitude of Resultsant field , | E | = { 9.743 × 9.743 + 1.125 × 1.125 }1/2  × 105  N/C
 
Magnitude of Resultsant field , | E |  = 9.808 × 105 N/C
 
Direction of field makes angle θ = tan-1 ( 1.125/9.743 ) = 6.6o counterclockwise from +x axis
 
Force experienced by α particle , F = qE = 3.2 × 10-19 × 9.808 × 105 = 3.138 × 10-13 N
Answered by Thiyagarajan K | 19 Sep, 2021, 03:22: PM

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