The electric field in a region is given by vecE = (Ax + B) hati where E is in NC-1and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is V1 and that at x = -5 is V2, then V1 - V2 (in volt) is:

Question

The electric field in a region is given by vecE = (Ax + B) hati where E is in NC-1and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x = 1 is V1 and that at x = -5 is V2, then V1 - V2 (in volt) is: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given, vecE=(A x+B) hati or E=20 x+10 Using V=∫ E d x, we have V2-V1=∫-51(20 x+10) d x=-180 V or V1-V2=180 V

Q. The electric field in a region is given by $E = (A x + B) \hat{i}$ where $E$ is in $N C^{- 1}$ and $x$ is in metres. The values of constants are $A = 20 S I$ unit and $B = 10 S I$ unit. If the potential at $x = 1$ is $V_{1}$ and that at $x = - 5$ is $V_{2}$ , then $V_{1} - V_{2}$ (in volt) is:

Given, $E = (A x + B) \hat{i}$
or $E = 20 x + 10$
Using $V = \int E d x$ , we have
$< b r / > V_{2} - V_{1} = \int_{- 5}^{1} (20 x + 10) d x = - 180 V < b r / >$
or $V_{1} - V_{2} = 180 V$

Q. The electric field in a region is given by E=(Ax+B)i^ where E is in NC−1and x is in metres. The values of constants are A=20SI unit and B=10SI unit. If the potential at x=1 is V1​ and that at x=−5 is V2​, then V1​−V2​ (in volt) is:

Given, E=(Ax+B)i^ or E=20x+10 Using V=∫Edx, we have <br/>V2​−V1​=∫−51​(20x+10)dx=−180V<br/> or V1​−V2​=180V

Q. The electric field in a region is given by $E = (A x + B) \hat{i}$ where $E$ is in $N C^{- 1}$ and $x$ is in metres. The values of constants are $A = 20 S I$ unit and $B = 10 S I$ unit. If the potential at $x = 1$ is $V_{1}$ and that at $x = - 5$ is $V_{2}$ , then $V_{1} - V_{2}$ (in volt) is:

Given, $E = (A x + B) \hat{i}$
or $E = 20 x + 10$
Using $V = \int E d x$ , we have
$< b r / > V_{2} - V_{1} = \int_{- 5}^{1} (20 x + 10) d x = - 180 V < b r / >$
or $V_{1} - V_{2} = 180 V$