Your resonance structure is perfectly alright. But the resonance structure with the negative and positive charges won't be very important and here is why: For one thing, it contains seperated charges and usually those resonance structures are most important that contain as few charges as possible. This is simply a matter of electrostatics - seperating charges and holding them apart from one another costs energy. A second reason has to do with the "electronic origin" of this resonance charge: It describes a conjugation between a free electron pair of $\ce{Br}$ and the $\pi$ system of the double bond - more accurately put the filled free-electron-pair-orbital can interact with the $\pi^{*}$ orbital of the $\ce{C=C}$ bond to form two new orbitals: an in-phase combination that is lower in energy and will contain the two electrons from the free electron pair and an out-of-phase combination that is higher in energy and will remain unoccupied. This stabilizes the system. The stronger the overlap between the free-electron-pair-orbital and the $\pi^{*}$ orbital of the $\ce{C=C}$ bond and the closer they are in energy the better the stabilization of the system through this conjugation is and the more important is the second resonance structure. But in the case at hand the $\pi$-overlap between the bromine orbital and the $\ce{C=C}$ bond is very weak (the reason for this you can find here) and so the conjugation is very weak, too. Thus, this resonance structure is not very important.