Two point charges q1 and q2 are placed at a distance 'd' apart as shown in the figure. The electric field intensity is zero at a point 'P' on the line joining them as shown. If the ratio q1q2 is −(xr+dr)2. Find the value of x.
Open in App
Solution
Verified by Toppr
Here, the electric field at P is EP=0
1) Since E=−dVdx, so dVPdx=0 . Thus the potential at P is maximum or minimum.
2) EP=kq2r2+kq1(r+d)2=0
⇒q1q2=−(r+dr)2
Was this answer helpful?
0
Similar Questions
Q1
Two point charges q1 and q2 are placed at a distance 'd' apart as shown in the figure. The electric field intensity is zero at a point 'P' on the line joining them as shown. If the ratio q1q2 is −(xr+dr)2. Find the value of x.
View Solution
Q2
Two point charges q1 and q2 are placed at a distance d apart as shown in the figure. The electric field intensity is zero at the point P on the line joining them as shown . Write two conclusions that you draw from this.
View Solution
Q3
Two points charges q1 and q2 are placed at a distance of a part as shown in the figure. The electric field intensity is zero at the point P on the line joining them as a shown. Write two conclusions that you can draw from this
View Solution
Q4
Charges +9Q and -4Q are placed as shown in figure. The point at which electric field intensity is zero at a distance from B on the line joining AB will be :
View Solution
Q5
Two charges +q and −q are placed at a distance b apart as shown in the figure. The electric field at a point P on the perpendicular bisector as shown is :