Get the information you need with the help of IDNLearn.com's extensive Q&A platform. Our community is ready to provide in-depth answers and practical solutions to any questions you may have.
Sagot :
The electric energy is 32 uj
The electric potential of a point charge is [tex]V=kQ/r[/tex], or [tex]V = k Q / r[/tex]. Electric potential is a scalar while the electric field is a vector. While the total electric field can be computed by adding individual fields as vectors, the voltage resulting from a collection of point charges can be calculated by adding voltages as integers.
It is decided to set the potential at infinity to zero. [tex]E=Fq=kQr^2[/tex], but V for a point charge reduces with distance while decreasing with distance squared. Remember that the electric field E is a vector, but the electric potential V is a scalar with no direction.
The electric energy is (8×4)=32 uj
Learn more about electric energy herebrainly.com/question/24403138
#4219.
The electric potential at that point in space is 8 v is 32uj
what is electric potential energy?
Potential energy is the energy within an object relative to its position and proximity to other objects within a field.
The electric potential of a point charge is V=kQ/rV=kQ/r , or V = k Q / rV=kQ/r . Electric potential is a scalar while the electric field is a vector. While the total electric field can be computed by adding individual fields as vectors, the voltage resulting from a collection of point charges can be calculated by adding voltages as integers.
It is decided to set the potential at infinity to zero.
but V for a point charge reduces with distance while decreasing with distance squared. Remember that the electric field E is a vector, but the electric potential V is a scalar with no direction.
The electric energy is (8×4)=32 uj
learn more about electric potential from here: https//brainly.com/question/28167853
#SPJ4
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com is your source for precise answers. Thank you for visiting, and we look forward to helping you again soon.
