CBSE Class 12-science Answered

derive coulombs law in vector form

Asked by Avadhut Katkar | 16 Aug, 2018, 12:56: PM

Expert Answer

Coulomb measured the force between two point charges and found that it varied inversly

as the square of the distance between the charges and was directly proportional to the product of the

magnitude of two charges and acted along the line of joining the two charges.

Thus if two point charges q1 and q2 are separated by a distance r in vacuum,

the magnitude of the force F between them is given by,

begin mathsize 12px style F space equals k space fraction numerator open vertical bar q subscript 1 cross times q subscript 2 close vertical bar over denominator r squared end fraction end style

....................(1)

The constant k in eqn.(1) is usually put as

begin mathsize 12px style k space equals space fraction numerator 1 over denominator 4 pi epsilon subscript 0 end fraction space end style

, so that Coulombs law is written as

begin mathsize 12px style F space equals fraction numerator 1 over denominator 4 pi epsilon subscript 0 end fraction space fraction numerator open vertical bar q subscript 1 cross times q subscript 2 close vertical bar over denominator r squared end fraction end style

.....................(2)

where ε_o is permitivity of free space and it is given by, ε_o= 8.854×10^-12 C²N^-1m^-2

Since force is vector, we need to write Coulombs law in vector notation.

(vector quantity is given in Bold)

Let the position vectors of charges q₁ and q₂ be r₁ and r₂ respectively. We denote force on q₁ due to q₂ by F₁₂ and

force on q₂ by q₁ by F₂₁ as shown in figure. The two point charges q₁ and q₂ have been numbered 1 and 2 for conveneience

and the vector leading from 1 to 2 is denoted by r₂₁

r₂₁ = r₂-r₁. The magnitude of vector r₂₁ is denoted by |r₂₁|. The direction of a vector is specified by unit vector along that vector.

We define the unit vector

begin mathsize 12px style bold r with bold hat on top subscript bold 21 bold space bold equals bold space fraction numerator bold r with bold rightwards arrow on top subscript bold 21 over denominator open vertical bar bold r with bold rightwards arrow on top subscript bold 21 close vertical bar end fraction end style

Coulombs force law between two point charges q₁ and q₂ located at r₁ and r₂ is then expressed as

begin mathsize 12px style bold F with bold rightwards arrow on top subscript bold 21 space equals space fraction numerator 1 over denominator 4 pi epsilon subscript 0 end fraction fraction numerator q subscript 1 q subscript 2 over denominator r subscript 21 superscript 2 end fraction space bold r with bold hat on top subscript bold 21 end style

The above equation is valid for any sign of q₁ and q₂. If q₁ and q₂ are of same sign, F₂₁ is along r₂₁, which denotes repulsion.

If they are opposite sign, F₂₁ is along -r₂₁ that denotes attraction. No need to write separate equation for like and unlike charges

Answered by Thiyagarajan K | 19 Aug, 2018, 02:22: PM

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