किसी विद्युत क्षेत्र में विद्युत विभव `V=K/r` द्वारा व्यक्त किया जाता है तो विद्युत क्षेत्र की तीव्रता ज्ञात कीजिए. यदि सदिश `vecr = 2 hati + 3 hatj + 6 hatk ` है।

Explanation

यदि V= V(x, y, 2) है तो
`E_x =- (del V)/(del x) , E_y = (-del V)/(del y) " तथा " E_z = (-del V)/(delz)`
तथा `vecE = E_x hati + E_y hatj + E_z hatk `
यहां `V = K/r = (K)/(sqrt(x^2 + y^2 + z^2) ) = K (x^2 + y^2 + z^2)^(-1//2)`
` E_x = -(del V)/(del x) = - (del)/(delx) [K(x^2 + y^2 + z^2 )^(-1//2) ]`
`[ because " यदि " vecr = xhati + y hatj + z hatk " तो " r = sqrt(x^2 +y^2 + z^2) ]`
`E_x = + 1/2 (K)/( (x^2 + y^2+ z^2 )^(3//2) ) .2x`
` E_x = + (Kx)/( (x^2 + y^2 + z^2)^(3//2) ) = + (Kx)/(r^3)`
इसी प्रकार ` E_y = - (del V)/(del y) = + (Ky)/(r^3)`
` E_z = (-delV)/(del z) = (+Kz)/(r^3)`
अतः ` vecE = (-del V)/(del x) hati - (delV)/(del y) hatj - (del V)/(del z) hatk`
` = (K)/(r^3)x hati + (K)/(r^3) y hatj + (K)/(r^3) hatk`
` (K)/(r^3) (x hati +y hatj + z hatk ) = (K)/(r^3) vec r`
` [ because " दिया है - " vecr = 2 hati + 3 hatj + 6 hatk `
` rArr r = sqrt(4+9+36) = sqrt(49) = 7`
` therefore r^3 = (7)^3 = 343] `
` vecE = (K)/(343) (2hati + 3 hatj + 6 hatk ) `