Question:

A cube of side 'a' has point charges \(+ Q\) located at each of its vertices except at the origin where the charge is \(-Q\). The electric field at the centre of cube is:
A cube of side 'a' has point charges +Q located at each of its vertices except at the origin where the charge is −Q

Updated On: Aug 21, 2024
  • $\frac{- Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })$
  • $\frac{-2 Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })$
  • $\frac{2 Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })$
  • $\frac{ Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

We can replace \(-Q\) charge at origin by \(+ Q\) and \(-2 Q .\) 
Now due to \(+Q\) charge at every corner of cube. Electric field at center of cube is zero so now net electric field at center is only due to \(-2 Q\) charge at origin. 
\(\vec{ E }=\frac{ kq \vec{ r }}{ r ^{3}}=\frac{1(-2 Q ) \frac{ a }{2}(\hat{ x }+\hat{ y }+\hat{ z })}{4 \pi \varepsilon_{0}\left(\frac{ a }{2} \sqrt{3}\right)^{3}}\)

\(\vec{ E }=\frac{-2 Q (\hat{ x }+\hat{ y }+\hat{ z })}{3 \sqrt{3} \pi a ^{2} \varepsilon_{0}}\)

Hence, The correct answer is option (B): \(\frac{-2 Q }{3 \sqrt{3} \pi \varepsilon_{0} a ^{2}}(\hat{ x }+\hat{ y }+\hat{ z })\)

Was this answer helpful?
0
0

Top Questions on Electric Field

View More Questions

Questions Asked in JEE Main exam

View More Questions

Concepts Used:

Electric Field

Electric Field is the electric force experienced by a unit charge. 

The electric force is calculated using the coulomb's law, whose formula is:

\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)

While substituting q2 as 1, electric field becomes:

 \(E=k\dfrac{|q_{1}|}{r^{2}}\)

SI unit of Electric Field is V/m (Volt per meter).