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Two point charges q_(A)=3muC and q_(B)=-3muC are located 0.2 m apart in vacuum. a. What is the electric field at the mid point O of the line AB joining the two charges? b. If a negative test charge of magnitude 1.5xx10^(-9)C is placed at this point, what is the force experienced by the test charge? |
Answer» Solution :(a) The situation is represented in the given figure . O is the MID - pointof line AB. Distance between the two charges , AB =20 cm`thereforeAO=OB=10cm` Net electric field at point O=E Electric field at point O caused by `+3muC` charge , `E_(1)=(3xx10^(-6))/(4piepsilon_(0)(AO)^(2))` `=(3xx10^(-6))/(4piepsilon_(0)(10xx10^(-2))^(2))N//C`alongOB where , `epsilon_(0)=`Permittivity of free space `(1)/(4piepsilon_(0))=9xx10^(9)Nm^(2)C^(-2)` Magnitude of electric field at point O caused by `-3muC` charge. `E_(2)=|(-3xx10^(-6))/(4piepsilon_(0)(OB)^(2))|` `=(3xx10^(-6))/(4piepsilon_(0)(10xx10^(-2))^(2))N//C` along OB `thereforeE=E_(1)+E_(2)` `=2xx(9xx10^(9))xx(-3xx10^(-6))/((10xx10^(-2))^(2))` [ Since the value of `E_(1)andE_(2)`are same , the value is multiplied with 2] `=5.4xx10^(6)N//C`along OB Therefore , the electric field at mid - point O is `5.4xx10^(6)NC^(-1)` along OB. (b) ATEST chrge of amount `1.5xx10^(-9)C` is PLACED at mid - point O. `q=1.5xx10^(-9)C` Force experienced by the test charge=F `thereforeF=qE` `=1.5xx10^(-9)xx5.4xx10^(6)` `=8.1xx10^(-3)N` 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 `8.1xx10^(-3)N` along OA. |
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