calculate the electric field due to the electric dipole of length 2 cm having charge 4 microcoulomb at an angle of minus 60 degree for the direction of electric dipole moment 1 meter for the center to the dipole

Asked by surajsinghrajput4691 | 16th May, 2022, 12:20: AM

Expert Answer:

Let the dipole is at origin of cartesian coordinate system as shown in figure.
 
electric field at P is given as
 
begin mathsize 14px style E space equals space fraction numerator q over denominator 4 pi epsilon subscript o end fraction open parentheses fraction numerator 1 over denominator r subscript plus superscript 2 end fraction space minus space fraction numerator 1 over denominator r subscript minus superscript 2 end fraction close parentheses end style
 
where q is magnitude of individual charge of dipole , r+ is ditsance between positive charge and point P , 
r- is ditsance between negative charge and point Pand εo is permittivity of free space .
 
distance r+ is determined as follows using cosine rule
 
begin mathsize 14px style r subscript plus superscript 2 space equals space r to the power of 2 to the power of space end exponent plus d squared space minus space 2 space r space d space cos 120 space equals space r to the power of 2 to the power of space end exponent plus d squared space plus space space r space d space
end style
where r is distance from origin to point P and 2d is length of dipole

begin mathsize 14px style r subscript plus superscript 2 space equals space r to the power of 2 to the power of space end exponent open parentheses 1 space plus stretchy left parenthesis d over r stretchy right parenthesis squared space plus space space d over r close parentheses space almost equal to r to the power of 2 to the power of space end exponent stretchy left parenthesis 1 space space plus space space d over r stretchy right parenthesis space
end style
begin mathsize 14px style fraction numerator 1 over denominator r subscript plus superscript 2 end fraction space equals space 1 over r squared space open parentheses 1 plus d over r close parentheses to the power of negative 1 end exponent space equals space 1 over r squared space open parentheses 1 minus d over r close parentheses end style
 
similarly we get
 
begin mathsize 14px style fraction numerator 1 over denominator r subscript minus superscript 2 end fraction space equals space 1 over r squared space open parentheses 1 plus d over r close parentheses end style
 
begin mathsize 14px style E space equals space fraction numerator negative 2 q over denominator 4 pi epsilon subscript o space r squared end fraction open parentheses d over r close parentheses space space equals space fraction numerator negative 2 q d over denominator 4 pi epsilon subscript o r cubed end fraction end style
let us substitute values
 
1/(4πεo ) = 9 × 109 N m2 C-2   , q = 4 × 10-6 C and d =1 cm
 
E = - ( 9 × 109 × 2 × 4 × 10-6 × 0.01 ) / 13 = -720 N

Answered by Thiyagarajan K | 16th May, 2022, 09:57: AM