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Q.
A cone lies in a uniform electric field $E$ as shown in figure. The electric flux entering the cone is
Electric Charges and Fields
Solution:
Plane normal to electric field is a triangle with base length $2 R$ and height $h$.
Area of triangle $A=\frac{1}{2} \times 2 R h=R h$
Electric flux entering the cone $=E A=E R h$
