For frist question I don't know the answer
bout I know for second question answer in attachment
Correct Answer - C
Electric field E = k q r 2 , E=kqr2, r = √ x 2 + y 2 + z 2 = √ 30 2 + 30 2 + 0 2 r=x2+y2+z2=302+302+02 =
Distance of point from origin = 30 √ 2 c m =302cm
So, |E|=9×109×0.008×10-6(302×10-2)2 = 400 N / C =400N/C
Also, unit vector in direction of (30, 30, 0) point from origin = ˆ i + ˆ j √ 2 =i^+j^2
So, E = Magnitude×direction = 400 × ( ˆ i + ˆ j √ 2 ) =400×(i^+j^2)
= 200 √ 2 ( ˆ i + ˆ ^j)
this is answer of 2 wues
Explanation:
im not getting the 1 ans
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For frist question I don't know the answer
bout I know for second question answer in attachment
Verified answer
Correct Answer - C
Electric field E = k q r 2 , E=kqr2, r = √ x 2 + y 2 + z 2 = √ 30 2 + 30 2 + 0 2 r=x2+y2+z2=302+302+02 =
Distance of point from origin = 30 √ 2 c m =302cm
So, |E|=9×109×0.008×10-6(302×10-2)2 = 400 N / C =400N/C
Also, unit vector in direction of (30, 30, 0) point from origin = ˆ i + ˆ j √ 2 =i^+j^2
So, E = Magnitude×direction = 400 × ( ˆ i + ˆ j √ 2 ) =400×(i^+j^2)
= 200 √ 2 ( ˆ i + ˆ ^j)
this is answer of 2 wues
Explanation:
im not getting the 1 ans