Q) Two point charges 3micro coulomb and -2micro coulomb are located at A and B which are 40 cm apart in vacuum.
a) What is the electric field at the midpoint O of the line joining the two charges?
b) If a negative test charge of magnitude 2.5 x10^-9°C is placed at this point, what is the force experienced by the test charge?
Answers
Explanation:
Solution
Solutionverified
SolutionverifiedVerified by Toppr
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram that
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′ ∩A=A
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′ ∩A=A(iii) It can be clearly seen from the Venn diagram , there is no common portion between A and A' .
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′ ∩A=A(iii) It can be clearly seen from the Venn diagram , there is no common portion between A and A' .Hence, A∩A
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′ ∩A=A(iii) It can be clearly seen from the Venn diagram , there is no common portion between A and A' .Hence, A∩A ′
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′ ∩A=A(iii) It can be clearly seen from the Venn diagram , there is no common portion between A and A' .Hence, A∩A ′ =ϕ
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′ ∩A=A(iii) It can be clearly seen from the Venn diagram , there is no common portion between A and A' .Hence, A∩A ′ =ϕ(iv) U
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′ ∩A=A(iii) It can be clearly seen from the Venn diagram , there is no common portion between A and A' .Hence, A∩A ′ =ϕ(iv) U ′
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′ ∩A=A(iii) It can be clearly seen from the Venn diagram , there is no common portion between A and A' .Hence, A∩A ′ =ϕ(iv) U ′ ∩A=ϕ∩A=ϕ
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′ ∩A=A(iii) It can be clearly seen from the Venn diagram , there is no common portion between A and A' .Hence, A∩A ′ =ϕ(iv) U ′ ∩A=ϕ∩A=ϕ∴U
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′ ∩A=A(iii) It can be clearly seen from the Venn diagram , there is no common portion between A and A' .Hence, A∩A ′ =ϕ(iv) U ′ ∩A=ϕ∩A=ϕ∴U ′
SolutionverifiedVerified by Toppr(i) It can be clearly seen from the Venn diagram thatA∪A ′ =U(ii) ϕ ′ ∩A=U∩A=A∴ϕ ′ ∩A=A(iii) It can be clearly seen from the Venn diagram , there is no common portion between A and A' .Hence, A∩A ′ =ϕ(iv) U ′ ∩A=ϕ∩A=ϕ∴U ′ ∩A=ϕ
Answer:
The situation is represented in the given figure. O is the mid-point of line AB. Distance between the two charges, AB = 20 cm ∴ AO = OB = 10 cm Net electric field at point O = E Electric field at point O caused by +3μC charge, Where, 0 = Permittivity of free space and 1/4πεo=9×109 Nm2C−2 Therefore, Magnitude of electric field at point O caused by −3μC charge, [ Since the magnitudes of E1 and E2 are equal and in the same direction ] Therefore, the electric field at mid-point O is 5.4 × 106 N C−1 along OB. (b) A test charge of amount 1.5 × 10−9 C is placed at mid – point O. q = 1.5 × 10−9 C Force experienced by the test charge = F The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A. Therefore, the force experienced by the test charge is 8.1 × 10−3 N along OA.Read more on Sarthaks.com - https://www.sarthaks.com/18169/two-point-charges-qa-3-c-and-qb-3-c-are-located-20-cm-apart-in-vacuum
Explanation: