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
| 19th Sep, 2021,
01:20: PM
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
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
Resultant field E due to both charge is wriiten as by consideing above fields in vector form
E = EA + EB = ( | EA | + | EB | ) cos30 
+ ( | EB | - | EA | ) sin30 
E = ( 9.743 + 1.125 ) × 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
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
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
Resultant field E due to both charge is wriiten as by consideing above fields in vector form
E = EA + EB = ( | EA | + | EB | ) cos30 + ( | EB | - | EA | ) sin30
+ ( | EB | - | EA | ) sin30 E = ( 9.743 + 1.125 ) × 105 N / C
+ 1.125 ) × 105 N / CMagnitude 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
| 19th Sep, 2021,
03:22: PM
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